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

sqrt(x^(2)-3x-6)=x-1
What extraneous solution does Tessa obtain?

x=

Tessa solves the equation below by first squaring both sides of the equation. $\newline$ $\sqrt{x^{2}-3 x-6}=x-1$ $\newline$ What extraneous solution does Tessa obtain? $\newline$ $x=$

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Evaluate recursive formulas for sequences

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

\newline

\sqrt{x^{2}-3 x-6}=x-1

\newline

What extraneous solution does Tessa obtain?

\newline

x=

Eliminate square root: Square both sides of the equation to eliminate the square root. $\newline$ $\sqrt{x^{2} - 3x - 6} = x - 1$ $\newline$ $(\sqrt{x^{2} - 3x - 6})^2 = (x - 1)^2$ $\newline$ $x^{2} - 3x - 6 = (x - 1)(x - 1)$
Expand right side: Expand the right side of the equation. $x^{2} - 3x - 6 = x^2 - 2x + 1$
Simplify equation: Subtract $x^2$ from both sides to simplify the equation. $\newline$ $x^{2} - x^2 - 3x - 6 = x^2 - x^2 - 2x + 1$ $\newline$ $-3x - 6 = -2x + 1$
Isolate x term: Add $2x$ to both sides to isolate the $x$ term on one side. $\newline$ $-3x + 2x - 6 = -2x + 2x + 1$ $\newline$ $-x - 6 = 1$
Solve for x: Add $6$ to both sides to solve for x. $\newline$ $-x - 6 + 6 = 1 + 6$ $\newline$ $-x = 7$
Check solution: Multiply both sides by $-1$ to find the value of $x$ . $-x \times -1 = 7 \times -1$ $x = -7$
Check solution: Multiply both sides by $-1$ to find the value of $x$ .
$-x \times -1 = 7 \times -1$
$x = -7$ Check the solution in the original equation to see if it is an extraneous solution.
$\sqrt{(-7)^2 - 3(-7) - 6} = -7 - 1$
$\sqrt{49 + 21 - 6} = -8$
$\sqrt{64} = -8$
$8 \neq -8$
The solution $x = -7$ does not satisfy the original equation, so it is an extraneous solution.

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