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Extraneous solutions of equations
Li Juan solves the equation below by first squaring both sides of the equation.

sqrt(3-2w)=w+6
What extraneous solution does Li Juan obtain?

w=

Extraneous solutions of equations $\newline$ Li Juan solves the equation below by first squaring both sides of the equation. $\newline$ $\sqrt{3-2 w}=w+6$ $\newline$ What extraneous solution does Li Juan obtain? $\newline$ $w=$

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Solve equations with variables on both sides

Full solution

Q. Extraneous solutions of equations

\newline

Li Juan solves the equation below by first squaring both sides of the equation.

\newline

\sqrt{3-2 w}=w+6

\newline

What extraneous solution does Li Juan obtain?

\newline

w=

Square Both Sides: Step $1$ : Square both sides of the equation to eliminate the square root. $\newline$ Original Equation: $\sqrt{3-2w} = w + 6$ $\newline$ Squared Equation: $(\sqrt{3-2w})^2 = (w + 6)^2$ $\newline$ $3 - 2w = w^2 + 12w + 36$
Rearrange and Set to Zero: Step $2$ : Rearrange the equation to set it to zero. $\newline$ $3 - 2w - w^2 - 12w - 36 = 0$ $\newline$ $-w^2 - 14w - 33 = 0$
Solve Quadratic Equation: Step $3$ : Solve the quadratic equation using the quadratic formula. $\newline$ $w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $\newline$ Here, $a = -1$ , $b = -14$ , $c = -33$ $\newline$ $w = \frac{-(-14) \pm \sqrt{(-14)^2 - 4(-1)(-33)}}{2(-1)}$ $\newline$ $w = \frac{14 \pm \sqrt{196 - 132}}{-2}$ $\newline$ $w = \frac{14 \pm \sqrt{64}}{-2}$ $\newline$ $w = \frac{14 \pm 8}{-2}$ $\newline$ $w = \frac{14 + 8}{-2}$ or $\frac{14 - 8}{-2}$ $\newline$ $a = -1$ $0$ or $a = -1$ $1$ $\newline$ $a = -1$ $2$ or $a = -1$ $3$
Check for Extraneous Solutions: Step $4$ : Check for extraneous solutions by substituting back into the original equation. $\newline$ Check $w = -11$ : $\newline$ $\sqrt{3 - 2(-11)} = -11 + 6$ $\newline$ $\sqrt{3 + 22} = -5$ $\newline$ $\sqrt{25} = -5$ $\newline$ $5 \neq -5$ (Extraneous)

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