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Rania solves the equation below by first squaring both sides of the equation.

-5=sqrt(3x-7)
What extraneous solution does Rania obtain?

x=

Rania solves the equation below by first squaring both sides of the equation. $\newline$ $-5=\sqrt{3 x-7}$ $\newline$ What extraneous solution does Rania obtain? $\newline$ $x=$

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Evaluate recursive formulas for sequences

Full solution

Q. Rania solves the equation below by first squaring both sides of the equation.

\newline

-5=\sqrt{3 x-7}

\newline

What extraneous solution does Rania obtain?

\newline

x=

Square both sides: Square both sides of the equation to eliminate the square root. $\newline$ $-5 = \sqrt{3x - 7}$ $\newline$ $(-5)^2 = (\sqrt{3x - 7})^2$ $\newline$ $25 = 3x - 7$
Add $7$ to isolate x: Add $7$ to both sides of the equation to isolate the term with x. $\newline$ $25 + 7 = 3x - 7 + 7$ $\newline$ $32 = 3x$
Divide by $3$ to solve for x: Divide both sides by $3$ to solve for x. $\newline$ $\frac{32}{3} = \frac{3x}{3}$ $\newline$ $x = \frac{32}{3}$
Check the solution: Check the solution by substituting $x$ back into the original equation. $\newline$ $-5 = \sqrt{3(\frac{32}{3}) - 7}$ $\newline$ $-5 = \sqrt{32 - 7}$ $\newline$ $-5 = \sqrt{25}$ $\newline$ $-5 = 5$ or $-5 = -5$ $\newline$ The original equation states $-5 = \sqrt{3x - 7}$ , so the square root should only yield a positive result, meaning $-5 = 5$ is not possible. Therefore, $x = \frac{32}{3}$ is an extraneous solution.

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