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Extraneous solutions of equations Rania solves the equation below by first squaring both sides of the equation. -5=sqrt(3x-7) What extraneous solution does Rania obtain? x= ◻

Extraneous solutions of equations\newlineRania solves the equation below by first squaring both sides of the equation.\newline−5=3x−7 -5=\sqrt{3 x-7} \newlineWhat extraneous solution does Rania obtain?\newlinex= x= \newline□ \square

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Q. Extraneous solutions of equations\newlineRania solves the equation below by first squaring both sides of the equation.\newline−5=3x−7 -5=\sqrt{3 x-7} \newlineWhat extraneous solution does Rania obtain?\newlinex= x= \newline□ \square
  1. Square Both Sides: Step 11: Square both sides of the equation to eliminate the square root.\newline−5=3x−7-5 = \sqrt{3x - 7}\newline(−5)2=(3x−7)2(-5)^2 = (\sqrt{3x - 7})^2\newline25=3x−725 = 3x - 7
  2. Solve for x: Step 22: Solve for x.\newline25=3x−725 = 3x - 7\newline25+7=3x25 + 7 = 3x\newline32=3x32 = 3x\newlinex=323x = \frac{32}{3}
  3. Check for Extraneous Solutions: Step 33: Check for extraneous solutions by substituting xx back into the original equation.x=323x = \frac{32}{3}−5=3(323)−7-5 = \sqrt{3(\frac{32}{3}) - 7}−5=32−7-5 = \sqrt{32 - 7}−5=25-5 = \sqrt{25}−5≠5-5 \neq 5

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