Learn on PengienVision, Algebra 2Chapter 5: Rational Exponents and Radical Functions

Lesson 4: Solving Radical Equations

In this Grade 11 enVision Algebra 2 lesson, students learn how to solve radical equations and inequalities by isolating the radical, raising both sides to an appropriate power, and checking for extraneous solutions. The lesson covers square root and cube root equations, rewriting radical formulas, and solving equations with rational exponents. Students practice identifying extraneous solutions that arise when both sides of an equation are squared or cubed during the solving process.

Section 1

Solving a Radical Equation

Property

A radical equation is one in which the variable appears under a square root or other radical. We solve simple radical equations by raising both sides to the appropriate power. To do this, first isolate the radical expression on one side of the equation. Then, raise both sides to the power that matches the index of the radical.

Examples

To solve $3\sqrt{x-4}=15$ , first divide by 3 to get $\sqrt{x-4}=5$ . Square both sides: $(\sqrt{x-4})^2 = 5^2$ , which gives $x-4=25$ , so $x=29$ .
To solve $\sqrt[3]{y+1}+6=9$ , first subtract 6 to get $\sqrt[3]{y+1}=3$ . Then cube both sides: $(\sqrt[3]{y+1})^3 = 3^3$ , which gives $y+1=27$ , so $y=26$ .
Solve $2\sqrt{3x-2}=10$ . Isolate the radical: $\sqrt{3x-2}=5$ . Square both sides: $3x-2=25$ . Solve for x: $3x=27$ , so $x=9$ .

Explanation

Think of this as unwrapping a present; raising to a power is the inverse operation that undoes a root. Isolating the radical first ensures that this unwrapping process is clean and doesn't create a more complicated expression to solve.

Section 2

Solving Radical Equations

Property

To Solve a Radical Equation

Isolate the radical on one side of the equation.
Square both sides of the equation.
Continue as usual to solve for the variable.

The technique of squaring both sides may introduce extraneous solutions, which are solutions that do not work in the original equation. Therefore, you must check your final answers.

Examples

To solve $\sqrt{x} = 8$ , square both sides: $(\sqrt{x})^2 = 8^2$ , which gives $x = 64$ . The check, $\sqrt{64}=8$ , works.
To solve $\sqrt{y-2} + 5 = 9$ , first isolate the radical to get $\sqrt{y-2} = 4$ . Squaring both sides gives $y-2 = 16$ , so $y = 18$ .
To solve $\sqrt{z} = -6$ , squaring both sides gives $z = 36$ . However, checking this in the original equation gives $\sqrt{36} = 6 \neq -6$ . Therefore, there is no solution.

Section 3

Equations with cube roots

Property

To solve an equation where the variable is under a cube root, first isolate the cube root. Then, undo the cube root by cubing both sides of the equation. We do not have to check for extraneous solutions when we cube both sides of an equation.

Examples

To solve $\sqrt[3]{y} = 4$ , we cube both sides: $(\sqrt[3]{y})^3 = 4^3$ , which gives the solution $y = 64$ .

Solve $2\sqrt[3]{x-1} = 6$ . First, isolate the radical by dividing by 2 to get $\sqrt[3]{x-1} = 3$ . Now, cube both sides: $(\sqrt[3]{x-1})^3 = 3^3$ , so $x-1=27$ , which gives $x=28$ .

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Solving a Radical Equation

Property

Examples

To solve $3\sqrt{x-4}=15$ , first divide by 3 to get $\sqrt{x-4}=5$ . Square both sides: $(\sqrt{x-4})^2 = 5^2$ , which gives $x-4=25$ , so $x=29$ .
To solve $\sqrt[3]{y+1}+6=9$ , first subtract 6 to get $\sqrt[3]{y+1}=3$ . Then cube both sides: $(\sqrt[3]{y+1})^3 = 3^3$ , which gives $y+1=27$ , so $y=26$ .
Solve $2\sqrt{3x-2}=10$ . Isolate the radical: $\sqrt{3x-2}=5$ . Square both sides: $3x-2=25$ . Solve for x: $3x=27$ , so $x=9$ .

Explanation

Section 2

Solving Radical Equations

Property

To Solve a Radical Equation

Isolate the radical on one side of the equation.
Square both sides of the equation.
Continue as usual to solve for the variable.

The technique of squaring both sides may introduce extraneous solutions, which are solutions that do not work in the original equation. Therefore, you must check your final answers.

Examples

To solve $\sqrt{x} = 8$ , square both sides: $(\sqrt{x})^2 = 8^2$ , which gives $x = 64$ . The check, $\sqrt{64}=8$ , works.
To solve $\sqrt{y-2} + 5 = 9$ , first isolate the radical to get $\sqrt{y-2} = 4$ . Squaring both sides gives $y-2 = 16$ , so $y = 18$ .
To solve $\sqrt{z} = -6$ , squaring both sides gives $z = 36$ . However, checking this in the original equation gives $\sqrt{36} = 6 \neq -6$ . Therefore, there is no solution.

Section 3

Equations with cube roots

Property

Examples

To solve $\sqrt[3]{y} = 4$ , we cube both sides: $(\sqrt[3]{y})^3 = 4^3$ , which gives the solution $y = 64$ .

Solve $2\sqrt[3]{x-1} = 6$ . First, isolate the radical by dividing by 2 to get $\sqrt[3]{x-1} = 3$ . Now, cube both sides: $(\sqrt[3]{x-1})^3 = 3^3$ , so $x-1=27$ , which gives $x=28$ .

Overview

Lesson overview

Review the lesson summary and core properties without leaving the current page.

Overview

Section 1

Solving a Radical Equation

Property

Examples

To solve $3\sqrt{x-4}=15$ , first divide by 3 to get $\sqrt{x-4}=5$ . Square both sides: $(\sqrt{x-4})^2 = 5^2$ , which gives $x-4=25$ , so $x=29$ .
To solve $\sqrt[3]{y+1}+6=9$ , first subtract 6 to get $\sqrt[3]{y+1}=3$ . Then cube both sides: $(\sqrt[3]{y+1})^3 = 3^3$ , which gives $y+1=27$ , so $y=26$ .
Solve $2\sqrt{3x-2}=10$ . Isolate the radical: $\sqrt{3x-2}=5$ . Square both sides: $3x-2=25$ . Solve for x: $3x=27$ , so $x=9$ .

Explanation

Section 2

Solving Radical Equations

Property

To Solve a Radical Equation

Isolate the radical on one side of the equation.
Square both sides of the equation.
Continue as usual to solve for the variable.

The technique of squaring both sides may introduce extraneous solutions, which are solutions that do not work in the original equation. Therefore, you must check your final answers.

Examples

To solve $\sqrt{x} = 8$ , square both sides: $(\sqrt{x})^2 = 8^2$ , which gives $x = 64$ . The check, $\sqrt{64}=8$ , works.
To solve $\sqrt{y-2} + 5 = 9$ , first isolate the radical to get $\sqrt{y-2} = 4$ . Squaring both sides gives $y-2 = 16$ , so $y = 18$ .
To solve $\sqrt{z} = -6$ , squaring both sides gives $z = 36$ . However, checking this in the original equation gives $\sqrt{36} = 6 \neq -6$ . Therefore, there is no solution.

Section 3

Equations with cube roots

Property

Examples

To solve $\sqrt[3]{y} = 4$ , we cube both sides: $(\sqrt[3]{y})^3 = 4^3$ , which gives the solution $y = 64$ .

Solve $2\sqrt[3]{x-1} = 6$ . First, isolate the radical by dividing by 2 to get $\sqrt[3]{x-1} = 3$ . Now, cube both sides: $(\sqrt[3]{x-1})^3 = 3^3$ , so $x-1=27$ , which gives $x=28$ .