Solving Radical Equations and Checking for Extraneous Solutions

Solving Radical Equations

Introduction

Radical equations are equations in which the variable is contained within a radical, most commonly a square root. Solving these equations often requires isolating the radical and then raising both sides of the equation to a power to eliminate the radical. It is important to check for extraneous solutions, which are solutions that arise from the process of squaring both sides but do not satisfy the original equation.

Key Steps for Solving Radical Equations

• Isolate the radical expression on one side of the equation if possible.

• Raise both sides to the appropriate power (usually square both sides for square roots) to eliminate the radical.

• Solve the resulting equation for the variable.

• Check all potential solutions in the original equation to ensure they are valid and not extraneous.

Examples and Solutions

Example 1: $\sqrt{2 - x} = x$

• Square both sides: $2 - x = x^2$

• Rearrange: $x^2 + x - 2 = 0$

• Factor: $(x + 2)(x - 1) = 0$

• Potential solutions: $x = -2,\ 1$

• Check $x = 1$: $\sqrt{2 - 1} = 1 \implies 1 = 1$ (valid)

• Check $x = -2$: $\sqrt{2 - (-2)} = -2 \implies \sqrt{4} = -2 \implies 2 = -2$ (not valid)

• Solution: $x = 1$

Example 2: $\sqrt{2x - 3} = 3 - x$

• Square both sides: $2x - 3 = (3 - x)^2$

• Expand: $2x - 3 = 9 - 6x + x^2$

• Rearrange: $x^2 - 8x + 12 = 0$

• Factor: $(x - 6)(x - 2) = 0$

• Potential solutions: $x = 6,\ 2$

• Check $x = 6$: $\sqrt{2(6) - 3} = 3 - 6 \implies \sqrt{12 - 3} = -3 \implies \sqrt{9} = -3 \implies 3 = -3$ (not valid)

• Check $x = 2$: $\sqrt{2(2) - 3} = 3 - 2 \implies \sqrt{4 - 3} = 1 \implies \sqrt{1} = 1 \implies 1 = 1$ (valid)

• Solution: $x = 2$

Example 3: $\sqrt{4b + 1} = -20$

• Note: The square root of a real number cannot be negative.

• Solution: No real solution.

Example 4: $\sqrt{5b} = \sqrt{3b + 2}$

• Square both sides: $5b = 3b + 2$

• Rearrange: $2b = 2 \implies b = 1$

• Check $b = 1$: $\sqrt{5(1)} = \sqrt{3(1) + 2} \implies \sqrt{5} = \sqrt{5}$ (valid)

• Solution: $b = 1$

Example 5: $4 + \sqrt{2n - 8} = n$

• Isolate the radical: $\sqrt{2n - 8} = n - 4$

• Square both sides: $2n - 8 = (n - 4)^2$

• Expand: $2n - 8 = n^2 - 8n + 16$

• Rearrange: $n^2 - 10n + 24 = 0$

• Factor: $(n - 4)(n - 6) = 0$

• Potential solutions: $n = 4,\ 6$

• Check $n = 4$: $4 + \sqrt{2(4) - 8} = 4 + \sqrt{8 - 8} = 4 + 0 = 4$ (valid)

• Check $n = 6$: $4 + \sqrt{2(6) - 8} = 4 + \sqrt{12 - 8} = 4 + 2 = 6$ (valid)

• Solution: $n = 4,\ 6$

Example 6: $14 = \sqrt{9x}$

• Square both sides: $196 = 9x$

• Solve: $x = \frac{196}{9}$

• Check: $14 = \sqrt{9 \cdot \frac{196}{9}} = \sqrt{196} = 14$ (valid)

• Solution: $x = \frac{196}{9}$

Checking for Extraneous Solutions

• Squaring both sides of an equation can introduce solutions that do not satisfy the original equation.

• Always substitute potential solutions back into the original equation to verify their validity.

• Extraneous solutions are common in radical equations, especially when the radical is set equal to a negative value.

Summary Table: Example Problems and Solutions

Equation	Potential Solutions	Valid Solutions
$\sqrt{2 - x} = x$	$x = -2,\ 1$	$x = 1$
$\sqrt{2x - 3} = 3 - x$	$x = 2,\ 6$	$x = 2$
$\sqrt{4b + 1} = -20$	None	None
$\sqrt{5b} = \sqrt{3b + 2}$	$b = 1$	$b = 1$
$4 + \sqrt{2n - 8} = n$	$n = 4,\ 6$	$n = 4,\ 6$
$14 = \sqrt{9x}$	$x = \frac{196}{9}$	$x = \frac{196}{9}$

Additional info:

• Some problems and solutions were inferred from partial or unclear handwriting, but all steps follow standard Precalculus methods for solving radical equations.