radicals problem solving with solution

8.6 Solve Radical Equations

Learning objectives.

By the end of this section, you will be able to:

• Solve radical equations
• Solve radical equations with two radicals
• Use radicals in applications

Be Prepared 8.16

Before you get started, take this readiness quiz.

Simplify: ( y − 3 ) 2 . ( y − 3 ) 2 . If you missed this problem, review Example 5.31 .

Be Prepared 8.17

Solve: 2 x − 5 = 0 . 2 x − 5 = 0 . If you missed this problem, review Example 2.2 .

Be Prepared 8.18

Solve n 2 − 6 n + 8 = 0 . n 2 − 6 n + 8 = 0 . If you missed this problem, review Example 6.45 .

Solve Radical Equations

In this section we will solve equations that have a variable in the radicand of a radical expression. An equation of this type is called a radical equation .

Radical Equation

An equation in which a variable is in the radicand of a radical expression is called a radical equation .

As usual, when solving these equations, what we do to one side of an equation we must do to the other side as well. Once we isolate the radical, our strategy will be to raise both sides of the equation to the power of the index. This will eliminate the radical.

Solving radical equations containing an even index by raising both sides to the power of the index may introduce an algebraic solution that would not be a solution to the original radical equation. Again, we call this an extraneous solution as we did when we solved rational equations.

In the next example, we will see how to solve a radical equation. Our strategy is based on raising a radical with index n to the n th power. This will eliminate the radical.

Example 8.56

How to solve a radical equation.

Solve: 5 n − 4 − 9 = 0 . 5 n − 4 − 9 = 0 .

Try It 8.111

Solve: 3 m + 2 − 5 = 0 . 3 m + 2 − 5 = 0 .

Try It 8.112

Solve: 10 z + 1 − 2 = 0 . 10 z + 1 − 2 = 0 .

Solve a radical equation with one radical.

• Step 1. Isolate the radical on one side of the equation.
• Step 2. Raise both sides of the equation to the power of the index.
• Step 3. Solve the new equation.
• Step 4. Check the answer in the original equation.

When we use a radical sign, it indicates the principal or positive root. If an equation has a radical with an even index equal to a negative number, that equation will have no solution.

Example 8.57

Solve: 9 k − 2 + 1 = 0 . 9 k − 2 + 1 = 0 .

Because the square root is equal to a negative number, the equation has no solution.

Try It 8.113

Solve: 2 r − 3 + 5 = 0 . 2 r − 3 + 5 = 0 .

Try It 8.114

Solve: 7 s − 3 + 2 = 0 . 7 s − 3 + 2 = 0 .

If one side of an equation with a square root is a binomial, we use the Product of Binomial Squares Pattern when we square it.

Binomial Squares

Don’t forget the middle term!

Example 8.58

Solve: p − 1 + 1 = p . p − 1 + 1 = p .

Try It 8.115

Solve: x − 2 + 2 = x . x − 2 + 2 = x .

Try It 8.116

Solve: y − 5 + 5 = y . y − 5 + 5 = y .

When the index of the radical is 3, we cube both sides to remove the radical.

Example 8.59

Solve: 5 x + 1 3 + 8 = 4 . 5 x + 1 3 + 8 = 4 .

Try It 8.117

Solve: 4 x − 3 3 + 8 = 5 4 x − 3 3 + 8 = 5

Try It 8.118

Solve: 6 x − 10 3 + 1 = −3 6 x − 10 3 + 1 = −3

Sometimes an equation will contain rational exponents instead of a radical. We use the same techniques to solve the equation as when we have a radical. We raise each side of the equation to the power of the denominator of the rational exponent. Since ( a m ) n = a m · n , ( a m ) n = a m · n , we have for example,

Remember, x 1 2 = x x 1 2 = x and x 1 3 = x 3 . x 1 3 = x 3 .

Example 8.60

Solve: ( 3 x − 2 ) 1 4 + 3 = 5 . ( 3 x − 2 ) 1 4 + 3 = 5 .

Try It 8.119

Solve: ( 9 x + 9 ) 1 4 − 2 = 1 . ( 9 x + 9 ) 1 4 − 2 = 1 .

Try It 8.120

Solve: ( 4 x − 8 ) 1 4 + 5 = 7 . ( 4 x − 8 ) 1 4 + 5 = 7 .

Sometimes the solution of a radical equation results in two algebraic solutions, but one of them may be an extraneous solution !

Example 8.61

Solve: r + 4 − r + 2 = 0 . r + 4 − r + 2 = 0 .

Try It 8.121

Solve: m + 9 − m + 3 = 0 . m + 9 − m + 3 = 0 .

Try It 8.122

Solve: n + 1 − n + 1 = 0 . n + 1 − n + 1 = 0 .

When there is a coefficient in front of the radical, we must raise it to the power of the index, too.

Example 8.62

Solve: 3 3 x − 5 − 8 = 4 . 3 3 x − 5 − 8 = 4 .

Try It 8.123

Solve: 2 4 a + 4 − 16 = 16 . 2 4 a + 4 − 16 = 16 .

Try It 8.124

Solve: 3 2 b + 3 − 25 = 50 . 3 2 b + 3 − 25 = 50 .

Solve Radical Equations with Two Radicals

If the radical equation has two radicals, we start out by isolating one of them. It often works out easiest to isolate the more complicated radical first.

In the next example, when one radical is isolated, the second radical is also isolated.

Example 8.63

Solve: 4 x − 3 3 = 3 x + 2 3 . 4 x − 3 3 = 3 x + 2 3 .

Try It 8.125

Solve: 5 x − 4 3 = 2 x + 5 3 . 5 x − 4 3 = 2 x + 5 3 .

Try It 8.126

Solve: 7 x + 1 3 = 2 x − 5 3 . 7 x + 1 3 = 2 x − 5 3 .

Sometimes after raising both sides of an equation to a power, we still have a variable inside a radical. When that happens, we repeat Step 1 and Step 2 of our procedure. We isolate the radical and raise both sides of the equation to the power of the index again.

Example 8.64

Solve: m + 1 = m + 9 . m + 1 = m + 9 .

Try It 8.127

Solve: 3 − x = x − 3 . 3 − x = x − 3 .

Try It 8.128

Solve: x + 2 = x + 16 . x + 2 = x + 16 .

We summarize the steps here. We have adjusted our previous steps to include more than one radical in the equation This procedure will now work for any radical equations.

Solve a radical equation.

• Step 1. Isolate one of the radical terms on one side of the equation.
• Step 3. Are there any more radicals? If yes, repeat Step 1 and Step 2 again. If no, solve the new equation.

Be careful as you square binomials in the next example. Remember the pattern is ( a + b ) 2 = a 2 + 2 a b + b 2 ( a + b ) 2 = a 2 + 2 a b + b 2 or ( a − b ) 2 = a 2 − 2 a b + b 2 . ( a − b ) 2 = a 2 − 2 a b + b 2 .

Example 8.65

Solve: q − 2 + 3 = 4 q + 1 . q − 2 + 3 = 4 q + 1 .

Try It 8.129

Solve: x − 1 + 2 = 2 x + 6 x − 1 + 2 = 2 x + 6

Try It 8.130

Solve: x + 2 = 3 x + 4 x + 2 = 3 x + 4

Use Radicals in Applications

As you progress through your college courses, you’ll encounter formulas that include radicals in many disciplines. We will modify our Problem Solving Strategy for Geometry Applications slightly to give us a plan for solving applications with formulas from any discipline.

Use a problem solving strategy for applications with formulas.

• Step 1. Read the problem and make sure all the words and ideas are understood. When appropriate, draw a figure and label it with the given information.
• Step 2. Identify what we are looking for.
• Step 3. Name what we are looking for by choosing a variable to represent it.
• Step 4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
• Step 5. Solve the equation using good algebra techniques.
• Step 6. Check the answer in the problem and make sure it makes sense.
• Step 7. Answer the question with a complete sentence.

One application of radicals has to do with the effect of gravity on falling objects. The formula allows us to determine how long it will take a fallen object to hit the gound.

Falling Objects

On Earth, if an object is dropped from a height of h feet, the time in seconds it will take to reach the ground is found by using the formula

For example, if an object is dropped from a height of 64 feet, we can find the time it takes to reach the ground by substituting h = 64 h = 64 into the formula.

It would take 2 seconds for an object dropped from a height of 64 feet to reach the ground.

Example 8.66

Marissa dropped her sunglasses from a bridge 400 feet above a river. Use the formula t = h 4 t = h 4 to find how many seconds it took for the sunglasses to reach the river.

Try It 8.131

A helicopter dropped a rescue package from a height of 1,296 feet. Use the formula t = h 4 t = h 4 to find how many seconds it took for the package to reach the ground.

Try It 8.132

A window washer dropped a squeegee from a platform 196 feet above the sidewalk Use the formula t = h 4 t = h 4 to find how many seconds it took for the squeegee to reach the sidewalk.

Police officers investigating car accidents measure the length of the skid marks on the pavement. Then they use square roots to determine the speed , in miles per hour, a car was going before applying the brakes.

Skid Marks and Speed of a Car

If the length of the skid marks is d feet, then the speed, s , of the car before the brakes were applied can be found by using the formula

Example 8.67

After a car accident, the skid marks for one car measured 190 feet. Use the formula s = 24 d s = 24 d to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

Try It 8.133

An accident investigator measured the skid marks of the car. The length of the skid marks was 76 feet. Use the formula s = 24 d s = 24 d to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

Try It 8.134

The skid marks of a vehicle involved in an accident were 122 feet long. Use the formula s = 24 d s = 24 d to find the speed of the vehicle before the brakes were applied. Round your answer to the nearest tenth.

Access these online resources for additional instruction and practice with solving radical equations.

• Solving an Equation Involving a Single Radical
• Solving Equations with Radicals and Rational Exponents
• Solving Radical Equations
• Radical Equation Application

Section 8.6 Exercises

Practice makes perfect.

In the following exercises, solve.

5 x − 6 = 8 5 x − 6 = 8

4 x − 3 = 7 4 x − 3 = 7

5 x + 1 = −3 5 x + 1 = −3

3 y − 4 = −2 3 y − 4 = −2

2 x 3 = −2 2 x 3 = −2

4 x − 1 3 = 3 4 x − 1 3 = 3

2 m − 3 − 5 = 0 2 m − 3 − 5 = 0

2 n − 1 − 3 = 0 2 n − 1 − 3 = 0

6 v − 2 − 10 = 0 6 v − 2 − 10 = 0

12 u + 1 − 11 = 0 12 u + 1 − 11 = 0

4 m + 2 + 2 = 6 4 m + 2 + 2 = 6

6 n + 1 + 4 = 8 6 n + 1 + 4 = 8

2 u − 3 + 2 = 0 2 u − 3 + 2 = 0

5 v − 2 + 5 = 0 5 v − 2 + 5 = 0

u − 3 + 3 = u u − 3 + 3 = u

v − 10 + 10 = v v − 10 + 10 = v

r − 1 = r − 1 r − 1 = r − 1

s − 8 = s − 8 s − 8 = s − 8

6 x + 4 3 = 4 6 x + 4 3 = 4

11 x + 4 3 = 5 11 x + 4 3 = 5

4 x + 5 3 − 2 = −5 4 x + 5 3 − 2 = −5

9 x − 1 3 − 1 = −5 9 x − 1 3 − 1 = −5

( 6 x + 1 ) 1 2 − 3 = 4 ( 6 x + 1 ) 1 2 − 3 = 4

( 3 x − 2 ) 1 2 + 1 = 6 ( 3 x − 2 ) 1 2 + 1 = 6

( 8 x + 5 ) 1 3 + 2 = −1 ( 8 x + 5 ) 1 3 + 2 = −1

( 12 x − 5 ) 1 3 + 8 = 3 ( 12 x − 5 ) 1 3 + 8 = 3

( 12 x − 3 ) 1 4 − 5 = −2 ( 12 x − 3 ) 1 4 − 5 = −2

( 5 x − 4 ) 1 4 + 7 = 9 ( 5 x − 4 ) 1 4 + 7 = 9

x + 1 − x + 1 = 0 x + 1 − x + 1 = 0

y + 4 − y + 2 = 0 y + 4 − y + 2 = 0

z + 100 − z = −10 z + 100 − z = −10

w + 25 − w = −5 w + 25 − w = −5

3 2 x − 3 − 20 = 7 3 2 x − 3 − 20 = 7

2 5 x + 1 − 8 = 0 2 5 x + 1 − 8 = 0

2 8 r + 1 − 8 = 2 2 8 r + 1 − 8 = 2

3 7 y + 1 − 10 = 8 3 7 y + 1 − 10 = 8

3 u + 7 = 5 u + 1 3 u + 7 = 5 u + 1

4 v + 1 = 3 v + 3 4 v + 1 = 3 v + 3

8 + 2 r = 3 r + 10 8 + 2 r = 3 r + 10

10 + 2 c = 4 c + 16 10 + 2 c = 4 c + 16

5 x − 1 3 = x + 3 3 5 x − 1 3 = x + 3 3

8 x − 5 3 = 3 x + 5 3 8 x − 5 3 = 3 x + 5 3

2 x 2 + 9 x − 18 3 = x 2 + 3 x − 2 3 2 x 2 + 9 x − 18 3 = x 2 + 3 x − 2 3

x 2 − x + 18 3 = 2 x 2 − 3 x − 6 3 x 2 − x + 18 3 = 2 x 2 − 3 x − 6 3

a + 2 = a + 4 a + 2 = a + 4

r + 6 = r + 8 r + 6 = r + 8

u + 1 = u + 4 u + 1 = u + 4

x + 1 = x + 2 x + 1 = x + 2

a + 5 − a = 1 a + 5 − a = 1

−2 = d − 20 − d −2 = d − 20 − d

2 x + 1 = 1 + x 2 x + 1 = 1 + x

3 x + 1 = 1 + 2 x − 1 3 x + 1 = 1 + 2 x − 1

2 x − 1 − x − 1 = 1 2 x − 1 − x − 1 = 1

x + 1 − x − 2 = 1 x + 1 − x − 2 = 1

x + 7 − x − 5 = 2 x + 7 − x − 5 = 2

x + 5 − x − 3 = 2 x + 5 − x − 3 = 2

In the following exercises, solve. Round approximations to one decimal place.

Landscaping Reed wants to have a square garden plot in his backyard. He has enough compost to cover an area of 75 square feet. Use the formula s = A s = A to find the length of each side of his garden. Round your answer to the nearest tenth of a foot.

Landscaping Vince wants to make a square patio in his yard. He has enough concrete to pave an area of 130 square feet. Use the formula s = A s = A to find the length of each side of his patio. Round your answer to the nearest tenth of a foot.

Gravity A hang glider dropped his cell phone from a height of 350 feet. Use the formula t = h 4 t = h 4 to find how many seconds it took for the cell phone to reach the ground.

Gravity A construction worker dropped a hammer while building the Grand Canyon skywalk, 4000 feet above the Colorado River. Use the formula t = h 4 t = h 4 to find how many seconds it took for the hammer to reach the river.

Accident investigation The skid marks for a car involved in an accident measured 216 feet. Use the formula s = 24 d s = 24 d to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

Accident investigation An accident investigator measured the skid marks of one of the vehicles involved in an accident. The length of the skid marks was 175 feet. Use the formula s = 24 d s = 24 d to find the speed of the vehicle before the brakes were applied. Round your answer to the nearest tenth.

Writing Exercises

Explain why an equation of the form x + 1 = 0 x + 1 = 0 has no solution.

ⓐ Solve the equation r + 4 − r + 2 = 0 . r + 4 − r + 2 = 0 . ⓑ Explain why one of the “solutions” that was found was not actually a solution to the equation.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction

• Authors: Lynn Marecek, Andrea Honeycutt Mathis
• Publisher/website: OpenStax
• Book title: Intermediate Algebra 2e
• Publication date: May 6, 2020
• Location: Houston, Texas
• Book URL: https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
• Section URL: https://openstax.org/books/intermediate-algebra-2e/pages/8-6-solve-radical-equations

© Jul 24, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

Radical Equations Practice Problems with Answers

We have eleven (11) radical equations lined up for you here. Each one is a bit different from the last. Dive in, and remember, it’s all about taking it one problem at a time. You’ve totally got this!

Problem 1: Solve the radical equation below.

[latex]\sqrt {2x \,- \,3} = – 1[/latex]

Since the square root of a real number must be positive , then this radical equation [latex]\color{red}\text{does not}[/latex] have a real solution. Notice the right-hand side is negative, that is, [latex]\color{red}-1[/latex].

Problem 2: Solve the radical equation below.

[latex]\sqrt {x + 9} = 4[/latex]

Start by squaring both sides of the equation. Then subtract both sides by [latex]9[/latex].

Verify the answer by substituting it back to the original radical equation to check if it yields a true statement.

• Check [latex]x={\color{red}7}[/latex]

Therefore, the solution is [latex]7[/latex].

Problem 3: Solve the radical equation below.

[latex]\sqrt { – 3x + 1} \,- \,5 = – 3[/latex]

Add [latex]5[/latex] to both sides of the equation. Square both sides to eliminate the square root. Subtract [latex]1[/latex] from both sides. Finally, divide both sides by [latex]-3[/latex].

Validate [latex]x=-1[/latex] from the original radical equation.

Therefore, [latex]x=-1[/latex] is a solution.

Problem 4: Solve the radical equation below.

[latex]\sqrt x \,- \,1 = 9\, – \,2x[/latex]

Start by isolating the square root term. Then, square both sides to eliminate the square root. Expand and simplify the resulting expression to form a quadratic equation. Factor out the trinomial to find the potential solutions. Finally, verify each solution by substituting it back into the original equation to ensure it is correct. The valid solution is identified through this verification process.

I will leave it to you verify [latex]x=4[/latex] and [latex]x ={\Large{ {25 \over 4}}}[/latex] by substituting each back to the original radical equation. You should find out that the only valid answer is [latex]x=4[/latex].

Problem 5: Solve the radical equation below.

[latex]\sqrt[3] { – 2x \,-\, 5} = – 3[/latex]

To solve the equation, first cube both sides to eliminate the cube root. Simplify the resulting equation and solve for [latex]x[/latex]. Verify the solution by substituting it back into the original equation to ensure it is correct. If it checks out, the solution is valid.

Therefore, [latex]x=11[/latex] is a solution to the given radical equation.

Problem 6: Solve the radical equation below.

[latex]\sqrt {2x + 1} \sqrt {3x \,- \,1} = 2[/latex]

To solve the radical equation, square both sides to eliminate the square roots. Multiply the binomials on the left side which resulting to quadratic equation. Simplify the terms, then use the factoring to find the solutions. Verify each solution in the original equation and identify the valid ones.

Note that the only solution is [latex]x ={\Large{ {5 \over 6}}}[/latex] which makes [latex]x=-1[/latex] an extraneous solution.

Problem 7: Solve the radical equation below.

[latex]\sqrt {4x\, – \,3} = 2x\, – \,9[/latex]

To solve the equation, first square both sides to eliminate the square root, then expand the right side. Move all terms to one side to form a quadratic equation and simplify it. Use the factoring method to solve for [latex]x[/latex]. Verify each solution by substituting it back into the original radical equation to check for validity. Determine which solutions satisfy the original equation and conclude with the valid one.

Here, [latex]x=3[/latex] is not a valid solution.

The only solution is [latex]x=7[/latex].

Problem 8: Solve the radical equation below.

[latex]\sqrt {2{x^2} \,- \,5x \,- \,7} = x + 1[/latex]

To solve the given equation, we first square both sides to eliminate the square root, then simplify the resulting equation. Next, we bring all terms to one side to set the equation to zero and solve the resulting quadratic equation using the factoring method . We then check each solution in the original equation to ensure they do not produce extraneous solutions. After verification, we determine that both solutions satisfy the original equation.

Thus, the solutions are [latex]x=8[/latex] and [latex]x=-1[/latex].

Problem 9: Solve the radical equation below.

[latex]\sqrt {3x + 1} = \sqrt {2x \,- \,1} + 1[/latex]

We start by squaring both sides of the equation to eliminate the square roots. After simplifying, we isolate the square root term once more on the right side of the equation. Then, we square both sides again to remove the square root term, resulting in a quadratic equation. We simplify this quadratic equation and proceed to solve it by factoring.

Lastly, we verify the obtained solutions by substituting them back into the original equation.

Therefore, the solutions are [latex]x=5[/latex] and [latex]x=1[/latex].

Problem 10: Solve the radical equation below.

[latex]\sqrt {x + 2} \,+ \,\sqrt {x\, – \,3} = 5[/latex]

Get rid of the square roots by squaring both sides of the equation. We will have to do it twice. Eventually, the quadratic term [latex]x^2[/latex] will be eliminated leaving a linear equation to solve which is very easy to address.

Verify that [latex]x=7[/latex] is a solution to the radical equation.

Problem 11: Solve the radical equation below.

[latex]\sqrt {2x \,- \,2} + \sqrt {2x + 7} = \sqrt {3x + 12}[/latex]

Square both sides of the equation. Simplify, then square both sides again to eliminate the square roots. Next, move all terms to the left side to set the right side equal to zero. Factor the trinomial on the left into two binomials. Lastly, set each binomial equal to zero to solve for [latex]x[/latex].

We need to verify our solutions from the original radical equation.

We should find out that [latex]x ={\Large{ {7 \over 5}}}[/latex] is true while [latex]x=-5[/latex] is false.

Therefore, the only solution is [latex]x ={\Large{ {7 \over 5}}}[/latex].

You might also like these tutorials:

• Radical Equations

Solving Radical Equations

How to solve equations with square roots, cube roots, etc.

Radical Equations

We can get rid of a square root by squaring (or cube roots by cubing, etc).

Warning: this can sometimes create "solutions" which don't actually work when we put them into the original equation. So we need to Check!

Follow these steps:

• isolate the square root on one side of the equation
• square both sides of the equation

Then continue with our solution!

Example: solve √(2x+9) − 5 = 0

Now it should be easier to solve!

Check: √(2·8+9) − 5 = √(25) − 5 = 5 − 5 = 0

That one worked perfectly.

More Than One Square Root

What if there are two or more square roots? Easy! Just repeat the process for each one.

It will take longer (lots more steps) ... but nothing too hard.

Example: solve √(2x−5) − √(x−1) = 1

We have removed one square root.

Now do the "square root" thing again:

We have now successfully removed both square roots.

Let us continue on with the solution.

It is a Quadratic Equation! So let us put it in standard form.

Using the Quadratic Formula (a=1, b=−14, c=29) gives the solutions:

2.53 and 11.47 (to 2 decimal places)

Let us check the solutions:

There is really only one solution :

Answer: 11.47 (to 2 decimal places)

See? This method can sometimes produce solutions that don't really work!

The root that seemed to work, but wasn't right when we checked it, is called an "Extraneous Root"

So checking is important.

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Unit 12: Radical equations & functions

About this unit.

This topic covers:

• Solving radical equations
• Graphing radical functions

Solving square-root equations

• Intro to square-root equations & extraneous solutions (Opens a modal)
• Intro to solving square-root equations (Opens a modal)
• Solving square-root equations (Opens a modal)
• Solving square-root equations: one solution (Opens a modal)
• Solving square-root equations: two solutions (Opens a modal)
• Solving square-root equations: no solution (Opens a modal)
• Square-root equations intro 4 questions Practice
• Square-root equations 4 questions Practice

Extraneous solutions of radical equations

• Square-root equations intro (Opens a modal)
• Equation that has a specific extraneous solution (Opens a modal)
• Extraneous solutions of radical equations (Opens a modal)
• Extraneous solutions of equations 4 questions Practice

Solving cube-root equations

• Solving cube-root equations (Opens a modal)

Domain of radical functions

• Domain of a radical function (Opens a modal)

Graphs of radical functions

• Transforming the square-root function (Opens a modal)
• Graphs of square-root functions (Opens a modal)
• Square-root functions & their graphs (Opens a modal)
• Radical functions & their graphs (Opens a modal)
• Graphs of square and cube root functions 4 questions Practice

Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices.

Roots and Radicals

Solve Radical Equations

Learning Objectives

By the end of this section, you will be able to:

• Solve radical equations
• Solve radical equations with two radicals
• Use radicals in applications

Before you get started, take this readiness quiz.

In this section we will solve equations that have a variable in the radicand of a radical expression. An equation of this type is called a radical equation .

An equation in which a variable is in the radicand of a radical expression is called a radical equation .

As usual, when solving these equations, what we do to one side of an equation we must do to the other side as well. Once we isolate the radical, our strategy will be to raise both sides of the equation to the power of the index. This will eliminate the radical.

Solving radical equations containing an even index by raising both sides to the power of the index may introduce an algebraic solution that would not be a solution to the original radical equation. Again, we call this an extraneous solution as we did when we solved rational equations.

In the next example, we will see how to solve a radical equation. Our strategy is based on raising a radical with index n to the n th power. This will eliminate the radical.

• Isolate the radical on one side of the equation.
• Raise both sides of the equation to the power of the index.
• Solve the new equation.
• Check the answer in the original equation.

When we use a radical sign, it indicates the principal or positive root. If an equation has a radical with an even index equal to a negative number, that equation will have no solution.

Because the square root is equal to a negative number, the equation has no solution.

If one side of an equation with a square root is a binomial, we use the Product of Binomial Squares Pattern when we square it.

Don’t forget the middle term!

When the index of the radical is 3, we cube both sides to remove the radical.

Sometimes the solution of a radical equation results in two algebraic solutions, but one of them may be an extraneous solution !

When there is a coefficient in front of the radical, we must raise it to the power of the index, too.

Solve Radical Equations with Two Radicals

If the radical equation has two radicals, we start out by isolating one of them. It often works out easiest to isolate the more complicated radical first.

In the next example, when one radical is isolated, the second radical is also isolated.

Sometimes after raising both sides of an equation to a power, we still have a variable inside a radical. When that happens, we repeat Step 1 and Step 2 of our procedure. We isolate the radical and raise both sides of the equation to the power of the index again.

We summarize the steps here. We have adjusted our previous steps to include more than one radical in the equation This procedure will now work for any radical equations.

• Isolate one of the radical terms on one side of the equation.

If yes, repeat Step 1 and Step 2 again.

Use Radicals in Applications

As you progress through your college courses, you’ll encounter formulas that include radicals in many disciplines. We will modify our Problem Solving Strategy for Geometry Applications slightly to give us a plan for solving applications with formulas from any discipline.

• Read the problem and make sure all the words and ideas are understood. When appropriate, draw a figure and label it with the given information.
• Identify what we are looking for.
• Name what we are looking for by choosing a variable to represent it.
• Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
• Solve the equation using good algebra techniques.
• Check the answer in the problem and make sure it makes sense.
• Answer the question with a complete sentence.

One application of radicals has to do with the effect of gravity on falling objects. The formula allows us to determine how long it will take a fallen object to hit the gound.

On Earth, if an object is dropped from a height of h feet, the time in seconds it will take to reach the ground is found by using the formula

It would take 2 seconds for an object dropped from a height of 64 feet to reach the ground.

Police officers investigating car accidents measure the length of the skid marks on the pavement. Then they use square roots to determine the speed , in miles per hour, a car was going before applying the brakes.

If the length of the skid marks is d feet, then the speed, s , of the car before the brakes were applied can be found by using the formula

Access these online resources for additional instruction and practice with solving radical equations.

• Solving an Equation Involving a Single Radical
• Solving Equations with Radicals and Rational Exponents
• Solving Radical Equations
• Radical Equation Application

Key Concepts

Practice Makes Perfect

In the following exercises, solve.

no solution

In the following exercises, solve. Round approximations to one decimal place.

Writing Exercises

Answers will vary.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

Intermediate Algebra Copyright © 2017 by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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How to Solve Radical Equations

Last Updated: March 11, 2023 Fact Checked

This article was co-authored by JohnK Wright V . JohnK Wright V is a Certified Math Teacher at Bridge Builder Academy in Plano, Texas. With over 20 years of teaching experience, he is a Texas SBEC Certified 8-12 Mathematics Teacher. He has taught in six different schools and has taught pre-algebra, algebra 1, geometry, algebra 2, pre-calculus, statistics, math reasoning, and math models with applications. He was a Mathematics Major at Southeastern Louisiana and he has a Bachelor of Science from The University of the State of New York (now Excelsior University) and a Master of Science in Computer Information Systems from Boston University. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 78,072 times.

Solving Equations with One Radical

• Final Answer: 64

• The expression above was expanded through Polynomial Multiplication. If you're confused how it was done, you can review the process here. [4] X Research source

Solving Equations with Multiple Radicals

• You frequently end up with quadratic equations when working with radicals. Review how to solve them if you are unsure.

• Using the quadratic equation, you only get two possible answers: 2.53 and 11.47.

Community Q&A

• You must check your solutions to end up at the right answers. Not all of the answers you find when solving radical equations are actual solutions. Thanks Helpful 0 Not Helpful 0
• The same steps work for cube roots, fourth roots and other roots. Instead of squaring both sides, for a cube root you will cube both sides. If you have a fourth root, you will raise both sides to the fourth power. This will work for any power. Thanks Helpful 0 Not Helpful 0

You Might Also Like

• ↑ https://www.mathsisfun.com/algebra/radical-equations-solving.html
• ↑ https://www.khanacademy.org/math/algebra2/radical-equations-and-functions/solving-square-root-equations/v/solving-radical-equations
• ↑ https://www.youtube.com/watch?v=jpD_BugTR6I
• ↑ https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratic-functions-equations/x2f8bb11595b61c86:untitled-1082/a/solving-quadratic-equations-by-taking-square-roots
• ↑ https://www.youtube.com/watch?v=gUh5ZNhI2lI
• ↑ https://www.youtube.com/watch?v=OSIxXgvrgC0

About This Article

To solve radical equations, which are any equations where the variable is under a square root, start by isolating the variable and radical on one side of the equation. Then, to undo the radical, square both sides of the equation. Check your answer by putting it back in the original equation. If the equation involves more complicated radicals, like the third root, follow the same procedure, but remove the radical by taking both sides of the equation to the third power. To learn how to use the same basic process for equations with multiple radicals, keep reading! Did this summary help you? Yes No

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Solving Equations with Radicals

A "radical" equation is an equation in which there is a variable inside the radical sign

Four steps to solve equations with radicals

Step 1: Isolate the radicals to left side of the equal sign.

Step 2: Square each side of the equation

Step 3: Solve the resulting equation

Step 4: Check all solutions

Equations with one radical

Example 1: Solve $\sqrt {2x + 3} - x = 0$

$$\sqrt {2x + 3} - x = 0$$ $$\sqrt {2x + 3} = x$$

$${\left( {\sqrt {2x + 3} } \right)^2} = {(x)^2}$$ $$2x + 2 = {x^2}$$ $${x^2} - 2x - 3 = 0$$

$${x^2} - 2x - 3 = 0$$ $$a = 1, b = - 2, c = - 3$$ $${x_{1,2}} = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$$ $${x_{1,2}} = \frac{{ - ( - 2) \pm \sqrt {{{( - 2)}^2} - 4 \cdot 1 \cdot ( - 3)} }}{{2 \cdot 1}}$$ $${x_{1,2}} = \frac{{2 \pm \sqrt {4 + 12} }}{2}$$ $${x_{1,2}} = \frac{{2 \pm 4}}{2}$$ $${x_1} = \frac{{2 + 4}}{2},{x_2} = \frac{{2 - 4}}{2}$$ $${x_1} = 3,{x_2} = - 1$$

Let's check to see if x 1 = 3 is solution:

$$\sqrt {2x + 3} - x = 0$$ $$\sqrt {2 \cdot 3 + 3} - 3 = 0$$ $$\sqrt 9 - 3 = 0$$ $$3 - 3 = 0$$ $$0 = 0, OK$$

Let's check to see if x 2 = -1 is solution:

$$\sqrt {2x + 3} - x = 0$$ $$\sqrt {2 \cdot ( - 1) + 3} - ( - 1) = 0$$ $$\sqrt { - 2 + 3} + 1 = 0$$ $$\sqrt 1 + 1 = 0$$ $$2 = 0,NOTOK$$

So, the original equation had a single solution x = 3 .

Exercise 1: Solve equations

Example 2: Solve $\sqrt {4x + 3} + 2x - 1 = 0$

$$\sqrt {4x + 3} + 2x - 1 = 0$$ $${(\sqrt {4x - 3} )^2} = {(1 - 2x)^2}$$ $$4x - 3 = 1 - 4x + 4{x^2}$$ $$4x - 3 - 1 + 4x - 4{x^2} = 0$$ $$ - 4{x^2} + 8x - 4 = 0/:( - 4)$$ $${x^2} - 2x + 1 = 0$$ $${(x - 1)^2} = 0$$ $$x = 1$$

Let's check it to see if x = 1 is a solution to the original equation.

$$\sqrt {4x - 3} + 2x - 1 = 0$$ $$\sqrt {4 \cdot 1 - 3} + 2 \cdot 1 - 1 = 0$$ $$\sqrt 1 + 2 - 1 = 0$$ $$2 = 0$$

So, the original equation had no solutions .

Exercise 2: Solve equations

Equations with two radicals

Example 3: Solve $\sqrt {3x + 4} - \sqrt {2x + 1} = 1$

First thing to do is get one of the square roots by itself.

$$\sqrt {3x + 4} - \sqrt {2x + 1} = 1$$ $$\sqrt {3x + 4} = 1 + \sqrt {2x + 1} $$ $${(\sqrt {3x + 4} )^2} = {(1 + \sqrt {2x + 1} )^2}$$ $$3x + 4 = {1^2} + 2 \cdot 1 \cdot \sqrt {2x + 1} + {(\sqrt {2x + 1} )^2}$$ $$3x + 4 = 1 + 2\sqrt {2x + 1} + 2x + 1$$

We have managed to eliminate one of square roots!! We will continue to work this problem as we did in the previous examples.

$$3x + 4 = 1 + 2\sqrt {2x + 1} + 2x + 1$$ $$3x + 4 = 2 + 2x + 2\sqrt {2x + 1} $$ $$3x + 4 - 2 - 2x = 2\sqrt {2x + 1} $$ $${(x + 2)^2} = {(2\sqrt {2x + 1} )^2}$$ $${x^2} + 2 \cdot 2 \cdot x + {2^2} = {2^2}(2x + 1)$$ $${x^2} + 4x + 4 = 8x + 4$$ $${x^2} - 4x = 0$$ $$x(x - 4) = 0$$ $${x_1} = 0$$ $${x_2} = 4$$

Let's check both possible solutions. We will start with x = 0 .

$$\sqrt {3x + 4} - \sqrt {2x + 1} = 1$$ $$\sqrt {3 \cdot 0 + 4} - \sqrt {2 \cdot 0 + 1} = 1$$ $$\sqrt 4 - \sqrt 1 = 1$$ $$1 = 1 \Rightarrow OK$$

Now let's check x = 4 .

$$\sqrt {3x + 4} - \sqrt {2x + 1} = 1$$ $$\sqrt {3 \cdot 4 + 4} - \sqrt {2 \cdot 4 + 1} = 1$$ $$\sqrt {16} - \sqrt 9 = 1$$ $$1 = 1 \Rightarrow OK$$

Exercise 3: Solve equations

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Solving Radical Equations

Learning Objective(s)

·          Solve equations containing radicals.

·          Recognize extraneous solutions.

·          Solve application problems that involve radical equations as part of the solution.

Introduction

An equation that contains a radical expression is called a radical equation . Solving radical equations requires applying the rules of exponents and following some basic algebraic principles. In some cases, it also requires looking out for errors generated by raising unknown quantities to an even power.

Squaring Both Sides

A basic strategy for solving radical equations is to isolate the radical term first, and then raise both sides of the equation to a power to remove the radical. (The reason for using powers will become clear in a moment.) This is the same type of strategy you used to solve other, non-radical equations—rearrange the expression to isolate the variable you want to know, and then solve the resulting equation.

Notice how you combined like terms and then squared both sides of the equation in this problem. This is a standard method for removing a radical from an equation. It is important to isolate a radical on one side of the equation and simplify as much as possible before squaring. The fewer terms there are before squaring, the fewer additional terms will be generated by the process of squaring.

In the example above, only the variable x was underneath the radical. Sometimes you will need to solve an equation that contains multiple terms underneath a radical. Follow the same steps to solve these, but pay attention to a critical point—square both sides of an equation, not individual terms . Watch how the next two problems are solved.

Extraneous Solutions

Following rules is important, but so is paying attention to the math in front of you—especially when solving radical equations. Take a look at this next problem that demonstrates a potential pitfall of squaring both sides to remove the radical.

Look at that—the answer a = 9 does not produce a true statement when substituted back into the original equation. What happened?

Incorrect values of the variable, such as those that are introduced as a result of the squaring process are called extraneous solutions . Extraneous solutions may look like the real solution, but you can identify them because they will not create a true statement when substituted back into the original equation. This is one of the reasons why checking your work is so important—if you do not check your answers by substituting them back into the original equation, you may be introducing extraneous solutions into the problem.

It may be difficult to understand why extraneous solutions exist at all. Thinking about extraneous solutions by graphing the equation may help you make sense of what is going on.

Although x = −1 is shown as a solution in both graphs, squaring both sides of the equation had the effect of adding an extraneous solution, x = −6. Again, this is why it is so important to check your answers when solving radical equations!

Solving Application Problems with Radical Equations

Radical equations play a significant role in science, engineering, and even music. Sometimes you may need to use what you know about radical equations to solve for different variables in these types of problems.

A common method for solving radical equations is to raise both sides of an equation to whatever power will eliminate the radical sign from the equation. But be careful—when both sides of an equation are raised to an even power, the possibility exists that extraneous solutions will be introduced. When solving a radical equation, it is important to always check your answer by substituting the value back into the original equation. If you get a true statement, then that value is a solution; if you get a false statement, then that value is not a solution.

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Solving Harder Radical Equations

Concepts Simple Eqns Harder Eqns Painful Eqns Higher-Index Eqns

Moving on from the simplest radical equations, we encounter equations in which the radical is not isolated. Sometimes, all we'll need to do is move a constant; other times, the solution will get rather messy.

Solve: katex.render("\\boldsymbol{\\color{green}{\\small{ \\sqrt{x\\,} - 2 = 5 }}}", typed01);

I could square both sides of the equation right away, but the radical isn't isolated on the left-hand side (because there's a 2 subtracted from it), so the result of squaring both sides at this stage won't be very helpful.

Here's what I get if I try squaring now:

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Solving Radical Equations

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I think I've made things worse! Rather than getting rid of the radical, I now have a radical term, plus a variable term outside of any radical. What now?

While squaring both sides at this point was not "wrong", it arguably wasn't much "right". (To complete the solution, I'd need to isolate the radical term, and then square both sides again.)

Instead of squaring right away, it will be more useful first to move the constant term 2 from the left-hand side of the equation over to the right-hand side. In this way, the radical will be isolated on the left:

I've isolated the square root, so now squaring both sides will work better:

Checking my solution, I find:

The solution checks, so my answer is:

Some radical equations can be simplified by first isolating the radical. But sometimes that won't be possible.

Find the solution: katex.render("\\boldsymbol{\\color{green}{\\small{ \\sqrt{x - 3\\,} - \\sqrt{x\\,} = 3 }}}", typed06);

This equation is a bit more messy than what we've seen before. I cannot isolate the radical because there are actually two radical terms. So how can I solve this algebraically? I'll have to square both sides twice . Here's what that looks like:

(sqrt[ x − 3] − sqrt[ x ]) 2 = (3) 2

( x − 3) − 2×sqrt[ x 2 − 3 x ] + ( x ) = 9

2 x − 12 = 2×sqrt[ x 2 − 3 x ]

x − 6 = sqrt[ x 2 − 3 x ]

At this point, having finishing squaring once, I have managed to get the radical isolated on the right-hand side of the equation. Now I'll square for the second time:

( x − 6) 2 = (sqrt[ x 2 − 3 x ]) 2

x 2 − 12 x + 36 = x 2 − 3 x     ⋆⋆

−9 x + 36 = 0

Checking my solution, I get:

LHS ≠ RHS

Hmm... All that work, and the only solution doesn't work in the original equation? Can that be right? I'll check the graph of the two lines corresponding to the two sides of the original equation to see if there appears to be an intersection point:

y 1 = sqrt[ x − 3] − sqrt[ x ]

According to the graph, no, it does not appear that these lines ever intersect (and calculus techniques can prove this). So my answer is:

no solution

Why did it appear that there was a solution to that equation? Let's look at the graphs after the second squaring of the radicals (starred above, and repeated below):

y 1 = x 2 − 12 x + 36

y 2 = x 2 − 3 x

By squaring (twice, in this case), I had accidentally created a solution that hadn't existed for the original equation. Because I checked that solution in the original equation, I was able to see that the actual answer was that the equation had no solution.

Because these exercises are so lengthy to solve, your book or instructor may not provide many examples. Try not to be misled by the small sample sizes. Yes, the above equation with two radicals had no solution. But, no, this will not always be the case for equations with two radicals.

Find the solution: katex.render("\\boldsymbol{\\color{green}{\\small{ \\sqrt{x - 3\\,} + \\sqrt{x\\,} = 3 }}}", typed30);

This is the same as the previous equation, except that the sign between the radicals has been reversed. If I turn the left-hand and right-hand sides of this equation into their own functions, I get:

Graphing them, I get:

So this equation does have a solution, which appears to be around x  = 4 . To know the answer for sure, I have to do the algebra:

...and here's the check of my solution:

Since the solution works in the original equation, then the solution is valid, and the answer is:

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Exponents and Radicals: Difficult Problems with Solutions

What Kamala Harris has said so far on key issues in her campaign

As she ramps up her nascent presidential campaign, Vice President Kamala Harris is revealing how she will address the key issues facing the nation.

In speeches and rallies, she has voiced support for continuing many of President Joe Biden’s measures, such as lowering drug costs , forgiving student loan debt and eliminating so-called junk fees. But Harris has made it clear that she has her own views on some key matters, particularly Israel’s treatment of Gazans in its war with Hamas.

In a departure from her presidential run in 2020, the Harris campaign has confirmed that she’s moved away from many of her more progressive stances, such as her interest in a single-payer health insurance system and a ban on fracking.

Harris is also expected to put her own stamp and style on matters ranging from abortion to the economy to immigration, as she aims to walk a fine line of taking credit for the administration’s accomplishments while not being jointly blamed by voters for its shortcomings.

Her early presidential campaign speeches have offered insights into her priorities, though she’s mainly voiced general talking points and has yet to release more nuanced plans. Like Biden, she intends to contrast her vision for America with that of former President Donald Trump. ( See Trump’s campaign promises here .)

“In this moment, I believe we face a choice between two different visions for our nation: one focused on the future, the other focused on the past,” she told members of the historically Black sorority Zeta Phi Beta at an event in Indianapolis in late July. “And with your support, I am fighting for our nation’s future.”

Here’s what we know about Harris’ views:

Harris took on the lead role of championing abortion rights for the administration after Roe v. Wade was overturned in June 2022. This past January, she started a “ reproductive freedoms tour ” to multiple states, including a stop in Minnesota thought to be the first by a sitting US president or vice president at an abortion clinic .

On abortion access, Harris embraced more progressive policies than Biden in the 2020 campaign, as a candidate criticizing his previous support for the Hyde Amendment , a measure that blocks federal funds from being used for most abortions.

Policy experts suggested that although Harris’ current policies on abortion and reproductive rights may not differ significantly from Biden’s, as a result of her national tour and her own focus on maternal health , she may be a stronger messenger.

High prices are a top concern for many Americans who are struggling to afford the cost of living after a spell of steep inflation. Many voters give Biden poor marks for his handling of the economy, and Harris may also face their wrath.

In her early campaign speeches, Harris has echoed many of the same themes as Biden, saying she wants to give Americans more opportunities to get ahead. She’s particularly concerned about making care – health care, child care, elder care and family leave – more affordable and available.

Harris promised at a late July rally to continue the Biden administration’s drive to eliminate so-called “junk fees” and to fully disclose all charges, such as for events, lodging and car rentals. In early August, the administration proposed a rule that would ban airlines from charging parents extra fees to have their kids sit next to them.

On day one, I will take on price gouging and bring down costs. We will ban more of those hidden fees and surprise late charges that banks and other companies use to pad their profits.”

Since becoming vice president, Harris has taken more moderate positions, but a look at her 2020 campaign promises reveals a more progressive bent than Biden.

As a senator and 2020 presidential candidate, Harris proposed providing middle-class and working families with a refundable tax credit of up to $6,000 a year (per couple) to help keep up with living expenses. Titled the LIFT the Middle Class Act, or Livable Incomes for Families Today, the measure would have cost at the time an estimated $3 trillion over 10 years.

Unlike a typical tax credit, the bill would allow taxpayers to receive the benefit – up to $500 – on a monthly basis so families don’t have to turn to payday loans with very high interest rates.

As a presidential candidate, Harris also advocated for raising the corporate income tax rate to 35%, where it was before the 2017 Tax Cuts and Jobs Act that Trump and congressional Republicans pushed through Congress reduced the rate to 21%. That’s higher than the 28% Biden has proposed.

Affordable housing was also on Harris’ radar. As a senator, she introduced the Rent Relief Act, which would establish a refundable tax credit for renters who annually spend more than 30% of their gross income on rent and utilities. The amount of the credit would range from 25% to 100% of the excess rent, depending on the renter’s income.

Harris called housing a human right and said in a 2019 news release on the bill that every American deserves to have basic security and dignity in their own home.

Consumer debt

Hefty debt loads, which weigh on people’s finances and hurt their ability to buy homes, get car loans or start small businesses, are also an area of interest to Harris.

As vice president, she has promoted the Biden administration’s initiatives on student debt, which have so far forgiven more than $168 billion for nearly 4.8 million borrowers . In mid-July, Harris said in a post on X that “nearly 950,000 public servants have benefitted” from student debt forgiveness, compared with only 7,000 when Biden was inaugurated.

A potential Harris administration could keep that momentum going – though some of Biden’s efforts have gotten tangled up in litigation, such as a program aimed at cutting monthly student loan payments for roughly 3 million borrowers enrolled in a repayment plan the administration implemented last year.

The vice president has also been a leader in the White House efforts to ban medical debt from credit reports, noting that those with medical debt are no less likely to repay a loan than those who don’t have unpaid medical bills.

In a late July statement praising North Carolina’s move to relieve the medical debt of about 2 million residents, Harris said that she is “committed to continuing to relieve the burden of medical debt and creating a future where every person has the opportunity to build wealth and thrive.”

Health care

Harris, who has had shifting stances on health care in the past, confirmed in late July through her campaign that she no longer supports a single-payer health care system .

During her 2020 campaign, Harris advocated for shifting the US to a government-backed health insurance system but stopped short of wanting to completely eliminate private insurance.

The measure called for transitioning to a Medicare-for-All-type system over 10 years but continuing to allow private insurance companies to offer Medicare plans.

The proposal would not have raised taxes on the middle class to pay for the coverage expansion. Instead, it would raise the needed funds by taxing Wall Street trades and transactions and changing the taxation of offshore corporate income.

When it comes to reducing drug costs, Harris previously proposed allowing the federal government to set “a fair price” for any drug sold at a cheaper price in any economically comparable country, including Canada, the United Kingdom, France, Japan or Australia. If manufacturers were found to be price gouging, the government could import their drugs from abroad or, in egregious cases, use its existing but never-used “march-in” authority to license a drug company’s patent to a rival that would produce the medication at a lower cost.

Harris has been a champion on climate and environmental justice for decades. As California’s attorney general, Harris sued big oil companies like BP and ConocoPhillips, and investigated Exxon Mobil for its role in climate change disinformation. While in the Senate, she sponsored the Green New Deal resolution.

During her 2020 campaign, she enthusiastically supported a ban on fracking — but a Harris campaign official said in late July that she no longer supports such a ban.

Fracking is the process of using liquid to free natural gas from rock formations – and the primary mode for extracting gas for energy in battleground Pennsylvania. During a September 2019 climate crisis town hall hosted by CNN, she said she would start “with what we can do on Day 1 around public lands.” She walked that back later when she became Biden’s running mate.

Biden has been the most pro-climate president in history, and climate advocates find Harris to be an exciting candidate in her own right. Democrats and climate activists are planning to campaign on the stark contrasts between Harris and Trump , who vowed to push America decisively back to fossil fuels, promising to unwind Biden’s climate and clean energy legacy and pull America out of its global climate commitments.

If elected, one of the biggest climate goals Harris would have to craft early in her administration is how much the US would reduce its climate pollution by 2035 – a requirement of the Paris climate agreement .

Immigration

Harris has quickly started trying to counter Trump’s attacks on her immigration record.

Her campaign released a video in late July citing Harris’ support for increasing the number of Border Patrol agents and Trump’s successful push to scuttle a bipartisan immigration deal that included some of the toughest border security measures in recent memory.

The vice president has changed her position on border control since her 2020 campaign, when she suggested that Democrats needed to “critically examine” the role of Immigration and Customs Enforcement, or ICE, after being asked whether she sided with those in the party arguing to abolish the department.

In June of this year, the White House announced a crackdown on asylum claims meant to continue reducing crossings at the US-Mexico border – a policy that Harris’ campaign manager, Julie Chavez Rodriguez, indicated in late July to CBS News would continue under a Harris administration.

Trump’s attacks stem from Biden having tasked Harris with overseeing diplomatic efforts in Central America in March 2021. While Harris focused on long-term fixes, the Department of Homeland Security remained responsible for overseeing border security.

She has only occasionally talked about her efforts as the situation along the US-Mexico border became a political vulnerability for Biden. But she put her own stamp on the administration’s efforts, engaging the private sector.

Harris pulled together the Partnership for Central America, which has acted as a liaison between companies and the US government. Her team and the partnership are closely coordinating on initiatives that have led to job creation in the region. Harris has also engaged directly with foreign leaders in the region.

Experts credit Harris’ ability to secure private-sector investments as her most visible action in the region to date but have cautioned about the long-term durability of those investments.

Israel-Hamas

The Israel-Hamas war is the most fraught foreign policy issue facing the country and has spurred a multitude of protests around the US since it began in October.

After meeting with Israeli Prime Minister Benjamin Netanyahu in late July, Harris gave a forceful and notable speech about the situation in Gaza.

We cannot look away in the face of these tragedies. We cannot allow ourselves to become numb to the suffering. And I will not be silent.”

Harris echoed Biden’s repeated comments about the “ironclad support” and “unwavering commitment” to Israel. The country has a right to defend itself, she said, while noting, “how it does so, matters.”

However, the empathy she expressed regarding the Palestinian plight and suffering was far more forceful than what Biden has said on the matter in recent months. Harris mentioned twice the “serious concern” she expressed to Netanyahu about the civilian deaths in Gaza, the humanitarian situation and destruction she called “catastrophic” and “devastating.”

She went on to describe “the images of dead children and desperate hungry people fleeing for safety, sometimes displaced for the second, third or fourth time.”

Harris emphasized the need to get the Israeli hostages back from Hamas captivity, naming the eight Israeli-American hostages – three of whom have been killed.

But when describing the ceasefire deal in the works, she didn’t highlight the hostage for prisoner exchange or aid to be let into Gaza. Instead, she singled out the fact that the deal stipulates the withdrawal by the Israeli military from populated areas in the first phase before withdrawing “entirely” from Gaza before “a permanent end to the hostilities.”

Harris didn’t preside over Netanyahu’s speech to Congress in late July, instead choosing to stick with a prescheduled trip to a sorority event in Indiana.

Harris is committed to supporting Ukraine in its fight against Russian aggression, having met with Ukrainian President Volodymyr Zelensky at least six times and announcing last month $1.5 billion for energy assistance, humanitarian needs and other aid for the war-torn country.

At the Munich Security Conference earlier this year, Harris said: “I will make clear President Joe Biden and I stand with Ukraine. In partnership with supportive, bipartisan majorities in both houses of the United States Congress, we will work to secure critical weapons and resources that Ukraine so badly needs. And let me be clear: The failure to do so would be a gift to Vladimir Putin.”

More broadly, NATO is central to our approach to global security. For President Biden and me, our sacred commitment to NATO remains ironclad. And I do believe, as I have said before, NATO is the greatest military alliance the world has ever known.”

Police funding

The Harris campaign has also walked back the “defund the police” sentiment that Harris voiced in 2020. What she meant is she supports being “tough and smart on crime,” Mitch Landrieu, national co-chair for the Harris campaign and former mayor of New Orleans, told CNN’s Pamela Brown in late July.

In the midst of nationwide 2020 protests sparked by George Floyd’s murder by a Minneapolis police officer, Harris voiced support for the “defund the police” movement, which argues for redirecting funds from law enforcement to social services. Throughout that summer, Harris supported the movement and called for demilitarizing police departments.

Democrats largely backed away from calls to defund the police after Republicans attempted to tie the movement to increases in crime during the 2022 midterm elections.

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Can immigration help solve canada’s recessionary risks.

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Yogi Yoganathan is CEO and Co-founder of RemitBee , a leading Canadian financial technology platform.

Immigration and its connotations have always made for lively discussion in Canada. Browse any media today, and it will likely yield some comments on immigration, with opinions falling on all sides of the aisle. These concerns are not all unfounded, given the gargantuan scale of immigration into the country. In 2024 alone, the government is set to admit 485,000 new planned permanent residents—an increase from the over 471,500 admissions last year.

To put this scale into perspective, immigration has replaced natural birth as the main driving force behind Canada's population boom , where immigrants or permanent residents account for nearly a quarter of the population .

While many were left scratching their heads at the necessity of such an aggressive immigration policy, it makes sense that the government would jump at an opportunity to offset the aging population, supplement its workforce and drive economic growth after a rough patch that included the Covid-19 pandemic.

The criticism has been free-flowing, from concerns of social friction to others linking the housing crisis and current recession-like state to increased immigration levels. While there's truth to these charges, I think the marketplace of ideas too often singles out immigration when discussing issues plaguing Canada. What if immigration was the solution, not the problem, to solving Canada's current recession risks?

International Students, The Answer To Canada's Labor Needs

Canada's workforce often requires a drastic input of skilled workers from outside sources. But what if the solution was already in our backyard? What if our institutions harbor reservoirs of skilled potential just waiting to be tapped and inducted into our workforce?

Canada had 1,040,985 international students across all levels at the end of 2023, of which 72.5% had plans to apply for a post-graduate work permit and 60% for permanent residency, according to the Canadian Bureau for International Education. The figures underline an interesting pipeline for the new generation of newcomers with established academic or work history in the country, Canadian degrees and a higher likelihood of having studied in STEM fields .

In addition to their professional contributions, the economy also benefits from having more residents paying taxes and banking within the system. Their increasing numbers and participation could play a vital role in pushing Canada out of its current slump.

Staying on the topic of highly skilled labor, Canada must also find a way to counter the poaching of its top university talents. Policies such as increasing the inclusion rate for capital gains tax from 50% to roughly 67% for gains exceeding $250,000 annually could push entrepreneurs to take their business elsewhere along with their talent.

Immigration And Banking

This brings us to the banking sector, which has witnessed its fair share of roadblocks from rate hikes, reduced money supply , prevailing geopolitical tensions, high levels of household debt and asset quality deterioration due to higher lending rates and inflation.

Given the prevailing conditions, a mass onboarding of new clients, including international students and workers, could provide a much-needed impetus for a banking sector under pressure and address the precarious positions of banks.

However, this path is not without its roadblocks. I mean specifically the regulations and formalities that stand between newcomers in Canada and access to credit. A newcomer here essentially starts from scratch when establishing their credit history and rating in the country, and while the steps are necessary to prevent defaults and ensure the health of the system, certain compromises and rehauling of the system to allow interoperability of credit scores across borders could go a long way in improving access and activity in the sector.

Curing A Recession

Before taking this any further, I'd like to point out that the suggestions for immigration as a solution to Canada's recession risks would be redundant without other complementary steps. Here's what I think are the most pertinent steps for instant results.

Increase Housing Supply

Canada must address its supply-side bottlenecks to ensure housing availability and affordability soon.

Fortunately, the government has been proactive in this regard, and its latest attempt, Canada's Housing Plan, which includes solutions such as low-cost loans to build apartments and removal of the goods and services tax for new rental apartment projects and co-ops, could help the economy get back on the right track. If it comes to fruition, the plan will increase the housing supply in the economy and help with the affordability of housing units—especially for a struggling middle class.

Improve Distribution Of Immigrant Talent

Uneven distribution is not alien to Canada, with most of its population concentrated within 100 miles of the American border. This issue also exists within the immigration system, where newcomers gravitate toward major metropolitan hubs such as Toronto, Vancouver and Montreal, seeking economic opportunities and familiarity with already existing immigrant communities.

This imbalance contributes to the pressure on housing and infrastructure within these hubs, inflating the market and perpetrating an affordability crisis among residents.

Create Interoperability Of Credit Scores Using Blockchain

A decentralized credit score with interoperability across borders would be a game-changer for the banking sector and the economy. Today, credit scores are determined by credit agencies and can often be influenced by bias or faulty data.

The technological leaps made through blockchain could be our way forward toward an interoperable and decentralized credit score system.

Parting Words

To sum it up, while it may seem counterintuitive, I believe that immigration done right could be the way out of this current lull. A larger influx of newcomers is a lesser threat than their exclusion from participating in the economy through strict banking regulations and antiquated credit scoring systems. The prevailing economic conditions certainly won't drum up any support or goodwill toward immigration, but demonizing it only prolongs the crisis without addressing the core issue.

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