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# Solving Radical Equations Practice Questions

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Updated on Sep 23, 2011

## Introduction

This set of practice questions will give you more practice operating with radicals. However, the focus here is to use radicals to solve equations. An equation is considered a radical equation when the radicand contains a variable. When you use a radical to solve an equation, you must be aware of the positive and negative roots. You should always check your results in the original equation to see that both solutions work. When one of the solutions does not work, it is called an extraneous solution. When neither solution works in the original equation, there is said to be no solution.

## Tips for Solving Radical Equations

• Squaring both sides of an equation is a valuable tool when solving radical equations. Use the following property: When a and b are algebraic expressions, if a = b, then a2 = b2.
• Isolate the radical on one side of an equation before using the squaring property.
• Squaring a radical results in the radical symbol disappearing, e.g., = x + 5.
• For second-degree equations, use the radical sign on both sides of the equation to find a solution for the variable. Check your answers.

## Practice Questions

Solve the following radical equations. Watch for extraneous solutions.

1. x2 = 49
2. x2 = 135
3. = 11
4. 2 = 24
5. – 4 = 4
6. = 8
7. + 8 = 12
8. + 3 = 12
9. 13
10. + 3 = 11
11. + 14 = 25
12. 3 = 15
13. 3 + 7 = 25
14. 3 = 10 –
15. + 24 = 38
16. 7 = 10 –
17. 3 – 4 = 29
18. = 7
19. x =
20. x =
21. x =
22. x =
23. = 2x
24. = x
25. x =

Numerical expressions in parentheses like this [ ] are operations performed on only part of the original expression. The operations performed within these symbols are intended to show how to evaluate the various terms that make up the entire expression.

Expressions with parentheses that look like this ( ) contain either numerical substitutions or expressions that are part of a numerical expression. Once a single number appears within these parentheses, the parentheses are no longer needed and need not be used the next time the entire expression is written.

When two pair of parentheses appear side by side like this ( )( ), it means that the expressions within are to be multiplied.

Sometimes parentheses appear within other parentheses in numerical or algebraic expressions. Regardless of what symbol is used, ( ), { }, or [ ], perform operations in the innermost parentheses first and work outward.

The solution is underlined.

 1. Use the radical sign on both sides of the equation. Show both solutions for the square root of 49. x = ±7 Check the first solution in the original equation. (7)2 = 49 49 = 49 Check the second solution in the original equation. (–7)2 = 49 49 = 49 Both solutions, check out.
 2. Use the radical sign on both sides of the equation. Simplify the radical. Check the first solution in the original equation. 9(15) = 135 135 = 135 Check the second solution in the original equation. (9)(15) = 135 135 = 135 Both solutions, check out.
3. First, square both sides of the equation.
Simplify both terms. n = 121
Check by substituting in the original equation.
The original equation asks for only the positive root of n. So when you substitute 121 into the original equation, only the positive root is to be considered. 11 = 11 checks out. Although this may seem trivial at this point, as the radical equations become more complex, this will become important.
 4. Isolate the radical on one side of the equation. Divide both sides by 2. Simplify terms. Now square both sides of the equation. Simplify terms. a = 144 Check the solution in the original equation. 2(12) = 24 The solution a = 144 checks out. 24 = 24
 5. Begin by adding 4 to both sides to isolate the radical. Combine like terms on each side. Square both sides of the equation. Simplify terms. 2x = 64 Divide both sides by 2. x = 32 Check the solution in the original equation. – 4 = 4 The solution x = 32 checks out. 8 – 4 = 4, 4 = 4
 6. Square both sides of the equation. 4x + 6 = 64 Subtract 6 from both sides of the equation. 4x = 58 Divide both sides by 4 and simplify. x = = 14.5 Check the solution in the original equation. Simplify terms. The solution x = 14.5 checks out. 8 = 8
 7. Subtract 8 from both sides in order to isolate the radical. = 4 Square both sides of the equation. Simplify terms. 3x + 4 = 16 Subtract 4 from both sides and divide by 3. x = 4 Check your solution in the original equation. Simplify terms. + 8 = 12 4 + 8 = 12 The solution x = 4 checks out. 12 = 12
 8. Subtract 3 from both sides of the equation isolating the radical. Square both sides of the equation. Simplify terms on both sides. 5x – 4 = 81 Add 4 to both sides and then divide by 5. x = = 17 Check your solution in the original equation. Simplify terms under the radical sign. Find the positive square root of 81. 9 + 3 = 12 Simplify. 12 = 12 The solution x = 17 checks out.
 9. Square both sides of the equation. Simplify terms on both sides of the equation. 4x + 9 = 169 Subtract 9 from both sides and then divide by 4. x = 40 Substitute the solution in the original equation. Simplify the expression under the radical sign. The radical sign calls for the positive square root. 13 ≠ –13 The solution does not check out. There is no solution for this equation.
 10. Subtract 3 from both sides isolating the radical. Square both sides of the equation. Simplify terms. 5x – 6 = 64 Add 6 to both sides and divide the result by 5. x = 14 Check the solution in the original equation. Simplify the expression under the radical. + 3 = 11 Find the positive square root of 64 and add 3. 8 + 3 = 11 The solution x = 14 checks out. 11 = 11
 11. Subtract 14 from both sides to isolate the radical. Now square both sides of the equation. 9 – x = 121 Subtract 9 from both sides. –x = 112 Multiply both sides by negative 1 to solve for x. x = –112 Check the solution in the original equation. Simplify the expression under the radical sign. The square root of 121 is 11. Add 14 and the solution checks. 25 = 25
 12. To isolate the radical, divide both sides by 3. Square both sides of the equation. Simplify terms. 3x + 1 = 25 Subtract 1 from both sides of the equation and divide by 3. x = 8 Check the solution in the original equation. Simplify the expression under the radical sign. Multiply 3 by the positive root of 25. 3(5) = 15 The solution x = 8 checks out. 15 = 15
 13. Subtract 7 from both sides of the equation. Divide both sides by 3 to isolate the radical. Square both sides of the equation. Simplify terms. –x = 36 Multiply both sides by negative 1. x = –36 Check the solution in the original equation. Simplify terms under the radical. Use the positive square root of 25. 3(6) + 7 = 25 18 + 7 = 25 The solution checks out. 25 = 25
 14.Add to both sides of the equation. Combine like terms and simplify the equation. 3 + = 10 Now subtract 3 from both sides. = 7 Square both sides of the equation. 100x – 1 = 49 Add 1 to both sides of the equation and divide by 100. x = 0.5 Check the solution in the original equation. Simplify the expression under the radical sign. The equation asks you to subtract the positive square root of 49 from 10. 3 = 10 – 7 The solution x = 0.5 checks out. 3 = 3
 15. Subtract 24 from both sides to isolate the radical. Square both sides of the equation. 3x + 46 = 196 Subtract 46 from both sides of the equation and divide by 3. x = 50 Check the solution in the original equation. Simplify the expression under the radical sign. 14 + 24 = 38 The solution x = 50 checks out. 38 = 38
 16. Add to both sides of the equation. – 7 = 10 – + Combine like terms and simplify the equation. – 7 = 10 Add 7 to both sides of the equation. = 17 Square both sides. 25x + 39 = 289 Subtract 39 from both sides and divide the result by 25. x = 10 Check the solution in the original equation. Simplify the expression under the radical sign. The equation asks you to subtract the positive square root of 289 from 10. –7 = 10 – 17 The solution x = 10 checks out. –7 = –7
 17. To isolate the radical on one side of the equation, add 4 to both sides and divide the result by 3. Square both sides of the equation. 13x + 43 = 121 Subtract 43 from both sides and divide by 13. x = 6 Check the solution in the original equation. Simplify the expression under the radical sign. Evaluate the left side of the equation. 3(11) – 4 = 29 The solution x = 6 checks out. 29 = 29
 18. To isolate the radical on one side of the equation, multiply both sides by . 28 = 7 Divide both sides of the equation by 7. 4 = Square both sides of the equation. 16 = 5x + 1 Subtract 1 from both sides and divide the result by 5. 3 = x Check the solution in the original equation. Simplify the expression under the radical sign. Divide the numerator by the positive square root of 16. The solution 3 = x checks out. 7 = 7
 19. The radical is alone on one side. Square both sides. x2 = 8 – 2x Transform the equation by putting all terms on one side. x2 + 2x – 8 = 0 The result is a quadratic equation. Solve for x by factoring using the trinomial factor form and setting each factor equal to zero and solving for x. (Refer to Chapter 16 for practice and tips for factoring quadratic equations.) It will be important to check each solution. x2 + 2x – 8 = (x + 4)(x – 2) = 0 Let the first factor equal zero and solve for x. x + 4 = 0 Subtract 4 from both sides. x = –4 Check the solution in the original equation. Evaluate the expression under the radical sign. The radical sign calls for a positive root. –4 ? 4 Therefore, x cannot equal –4. x = –4 is an example of an extraneous root. Let the second factor equal zero and solve for x. x – 2 = 0 Subtract 2 from both sides. x = 2 Check the solution in the original equation. Evaluate the expression under the radical sign. The positive square root of 4 is 2. 2 = 2 Therefore, the only solution for the equation is x = 2.
 20. With the radical alone on one side of the equation, square both sides. x2 = 3x + 4 The resulting quadratic equation may have up to two solutions. Put it into standard form and factor the equation using the trinomial factor form to find the solutions. Then check the solutions in the original equation. x2 – 3x – 4 = (x – 4)(x + 1) = 0 Letting each factor equal zero and solving for x results in two possible solutions, x = 4 and/or –1. Check the first possible solution in the original equation. The solution checks out. Now check the second possible solution in the original equation. –1 ≠ 1 Therefore, x ≠ –1. x = –1 is an extraneous root. The only solution for the original equation is x = 4.
 21. Square both sides of the equation. x2 = x + 12 Subtract (x + 12) from both sides of the equation. x2 – x – 12 = 0 The resulting quadratic equation may have up to two solutions. Factor the equation to find the solutions and check in the original equation. x2 – x – 12 = (x – 4)(x + 3) = 0 Let the first factor equal zero and solve for x. x – 4 = 0, so x = 4 Let the second factor equal zero and solve for x. x + 3 = 0, so x = –3 Check each solution in the original equation to rule out an extraneous solution. Simplify the expression under the radical sign. The solution x = 4 checks out. Now check the second possible solution in the original equation. You could simplify the expression under the radical sign to get the square root of 9. However, the radical sign indicates that the positive solution is called for, and the left side of the original equation when x = –3 is a negative number. So, x = –3 is not a solution.
 22. Square both sides of the equation. x2 = 7x – 10 Subtract (7x – 10) from both sides of the equation. x2 – 7x + 10 = 0 Factor the quadratic equation to find the solutions, and check each in the original equation to rule out any extraneous solution. x2 – 7x + 10 = (x – 5)(x – 2) = 0 The first factor will give you the solution x = 5. The second factor will give the solution x = 2. Check the first solution for the quadratic equation in the original equation. Simplify the expression under the radical sign. The solution x = 5 is a solution to the original equation. Now check the second solution to the quadratic equation in the original. Simplify the expression under the radical sign. There are two solutions to the original equation, x = 2 and x = 5
 23. Square both sides of the radical equation. 4x + 3 = 4x2 Transform the equation into a quadratic equation. 4x2 – 4x – 3 = 0 Factor the result using the trinomial factor form. 4x2 – 4x – 3 = (2x + 1)(2x – 3) = 0 Let the first factor equal zero and solve for x. 2x + 1 = 0, so x = –0.5 Let the second factor equal zero and solve for x. 2x – 3 = 0, so x = 1.5 When you substitute –0.5 for x in the original equation, the result will be = –1. That cannot be true for the original equation, so x ≠ –0.5. Substitute 1.5 for x in the original equation. Simplify the terms on each side of the equal sign. The only solution for the original equation is x = 1.5.
 24. Square both sides of the equation. 2 – = x2 Subtract (2 – ) from both sides of the equation. 0 = x2 + – 2 Multiply both sides of the equation by 2 to simplify the fraction. 0 = 2x2 + 7x – 4 Factor using the trinomial factor form. 2x2 + 7x – 4 = (2x – 1)(x + 4) = 0 Let the first factor equal zero and solve for x. 2x – 1 = 0, so x = Let the second factor equal zero and solve for x. x + 4 = 0, so x = –4 Check the first possible solution in the original equation. Simplify the expression under the radical sign. = So x = is a solution. Check the solution x = –4 in the original equation. Simplify the expression. This is not true. There is one solution for the original equation, x = .
 25. Square both sides of the equation. x2 = + 10 Add ( – 10) to both sides of the equation. x2 = – 10 = 0 Multiply the equation by 2 to eliminate the fraction. 2x2 – 3x – 20 = 0 Factor using the trinomial factor form. 2x2 – 3x – 20 = (x – 4)(2x + 5) = 0 Letting each factor of the trinomial factors equal zero results in two possible solutions for the original equation, x = 4 and/or x = . Check the first possible solution in the original equation. Simplify the radical expression. The solution x = 4 checks out as a solution for the original equation. Check the second possible solution in the original equation. A negative number cannot be equal to a positive square root as the radical sign in the original expression calls for. Therefore, x = is not a solution to the original equation. The only solution for this equation is x = 4.

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