Full solution

Q. 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?

Square both sides: Square both sides of the equation. $\newline$ To eliminate the square root, we square both sides of the equation $\sqrt{3-2w}=w+6$ . $\newline$ $(\sqrt{3-2w})^2 = (w+6)^2$ $\newline$ $3-2w = w^2 + 12w + 36$
Rearrange to zero: Rearrange the equation to set it to zero. $\newline$ To solve for $w$ , we need to rearrange the equation into a standard quadratic form. $\newline$ $0 = w^2 + 12w + 36 + 2w - 3$ $\newline$ $0 = w^2 + 14w + 33$
Factor quadratic equation: Factor the quadratic equation. $\newline$ We look for two numbers that multiply to $33$ and add up to $14$ . These numbers are $11$ and $3$ . $\newline$ $0 = (w + 11)(w + 3)$
Solve for w: Solve for $w$ . $\newline$ Set each factor equal to zero and solve for $w$ . $\newline$ $w + 11 = 0$ or $w + 3 = 0$ $\newline$ $w = -11$ or $w = -3$
Check extraneous solutions: Check for extraneous solutions. $\newline$ We need to check both solutions in the original equation $\sqrt{3-2w}=w+6$ to see if they produce any extraneous solutions. $\newline$ First, check $w = -11$ : $\newline$ $\sqrt{3-2(-11)} = -11 + 6$ $\newline$ $\sqrt{3+22} = -5$ $\newline$ $\sqrt{25} = -5$ $\newline$ This is not true because the square root of a positive number cannot be negative.
Check extraneous solutions: Check for extraneous solutions. $\newline$ We need to check both solutions in the original equation $\sqrt{3-2w}=w+6$ to see if they produce any extraneous solutions. $\newline$ First, check $w = -11$ : $\newline$ $\sqrt{3-2(-11)} = -11 + 6$ $\newline$ $\sqrt{3+22} = -5$ $\newline$ $\sqrt{25} = -5$ $\newline$ This is not true because the square root of a positive number cannot be negative.Check the second solution $w = -3$ : $\newline$ $\sqrt{3-2(-3)} = -3 + 6$ $\newline$ $\sqrt{3+6} = 3$ $\newline$ $\sqrt{9} = 3$ $\newline$ This is true because the square root of $9$ is indeed $3$ .