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=

Square both sides: Square both sides of the equation to eliminate the square root. (sqrt(3 - 2w))^2 = (w + 6)^2 3 - 2w = w^2 + 12w + 36Rearrange and find values: Rearrange the equation to set it to zero and find the values of w. 0 = w^2 + 12w + 36 - (3 - 2w) 0 = w^2 + 12w + 36 - 3 + 2w 0 = w^2 + 14w + 33Factor the quadratic: Factor the quadratic equation. (w + 11)(w + 3) = 0Solve for w: Solve for w by setting each factor equal to zero. w + 11 = 0 or w + 3 = 0 w = -11 or w = -3Check for extraneous solutions: Check for extraneous solutions by substituting the values of w back into the original equation. Check w = -11: sqrt(3 - 2(-11)) = -11 + 6 sqrt(3 + 22) = -5 sqrt(25) = -5 5 ≠ -5 (This is an extraneous solution because the square root of a positive number cannot be negative.)

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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=

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 $\mathrm{Li}$ Juan obtain? $\newline$ $w=$

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Q. 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

\mathrm{Li}

Juan obtain?

\newline

w=

Square both sides: Square both sides of the equation to eliminate the square root. $\newline$ $(\sqrt{3 - 2w})^2 = (w + 6)^2$ $\newline$ $3 - 2w = w^2 + 12w + 36$
Rearrange and find values: Rearrange the equation to set it to zero and find the values of $w$ . $0 = w^2 + 12w + 36 - (3 - 2w)$ $0 = w^2 + 12w + 36 - 3 + 2w$ $0 = w^2 + 14w + 33$
Factor the quadratic: Factor the quadratic equation. $\newline$ $(w + 11)(w + 3) = 0$
Solve for $w$ : Solve for $w$ by setting each factor equal to zero. $w + 11 = 0$ or $w + 3 = 0$ $w = -11$ or $w = -3$
Check for extraneous solutions: Check for extraneous solutions by substituting the values of $w$ back into the original equation. $\newline$ Check $w = -11$ : $\newline$ $\sqrt{3 - 2(-11)} = -11 + 6$ $\newline$ $\sqrt{3 + 22} = -5$ $\newline$ $\sqrt{25} = -5$ $\newline$ $5 \neq -5$ (This is an extraneous solution because the square root of a positive number cannot be negative.)