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

sqrt(z^(2)+8)=1-2z
What extraneous solution does Avery obtain?

z=

Avery solves the equation below by first squaring both sides of the equation. $\newline$ $\sqrt{z^{2}+8}=1-2 z$ $\newline$ What extraneous solution does Avery obtain? $\newline$ $z=$

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Evaluate recursive formulas for sequences

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

\newline

\sqrt{z^{2}+8}=1-2 z

\newline

What extraneous solution does Avery obtain?

\newline

z=

Rephrasing the problem: First, let's rephrase the ＂What extraneous solution does Avery obtain when solving the equation involving a square root by squaring both sides?＂
Squaring both sides: Avery starts with the equation $\sqrt{z^2 + 8} = 1 - 2z$ . To eliminate the square root, Avery squares both sides of the equation. $\newline$ (\sqrt{z^\(2\) + \(8\)})^\(2 = ( $1$ - $2$ z)^ $2$
Expanding the equation: After squaring both sides, the equation becomes: $z^2 + 8 = (1 - 2z)^2$
Rearranging the equation: Next, Avery expands the right side of the equation: $\newline$ $z^2 + 8 = 1^2 - 2(1)(2z) + (2z)^2$ $\newline$ $z^2 + 8 = 1 - 4z + 4z^2$
Combining like terms: Now, Avery rearranges the equation to set it to zero: $\newline$ $0 = 4z^2 - 4z + 1 - z^2 - 8$
Using the quadratic formula: Avery combines like terms: $0 = 3z^2 - 4z - 7$
Simplifying the square root: Avery uses the quadratic formula to solve for $z$ :
$z = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(-7)}}{2 \cdot 3}$
$z = \frac{4 \pm \sqrt{16 + 84}}{6}$
$z = \frac{4 \pm \sqrt{100}}{6}$
Finding the solutions: Avery simplifies the square root and continues solving: $\newline$ $z = \frac{4 \pm 10}{6}$ $\newline$ This gives two solutions: $z = \frac{(4 + 10)}{6}$ and $z = \frac{(4 - 10)}{6}$
Checking the solutions: Avery finds the two solutions: $\newline$ z = $\frac{14}{6}$ and z = $\frac{-6}{6}$ $\newline$ z = $\frac{7}{3}$ and z = $-1$
Checking the solutions: Avery finds the two solutions: $\newline$ $z = \frac{14}{6}$ and $z = \frac{-6}{6}$ $\newline$ $z = \frac{7}{3}$ and $z = -1$ Avery checks both solutions by substituting them back into the original equation: $\newline$ For $z = \frac{7}{3}$ : $\newline$ $\sqrt{\left(\frac{7}{3}\right)^2 + 8} = 1 - 2\left(\frac{7}{3}\right)$ $\newline$ $\sqrt{\frac{49}{9} + \frac{72}{9}} = 1 - \frac{14}{3}$ $\newline$ $\sqrt{\frac{121}{9}} = 1 - \frac{14}{3}$ $\newline$ $\frac{11}{3} = 1 - \frac{14}{3}$ $\newline$ $\frac{11}{3} = -\frac{11}{3}$ $\newline$ This is not true, so $z = \frac{7}{3}$ is an extraneous solution.
Checking the solutions: Avery finds the two solutions: $\newline$ $z = \frac{14}{6}$ and $z = \frac{-6}{6}$ $\newline$ $z = \frac{7}{3}$ and $z = -1$ Avery checks both solutions by substituting them back into the original equation: $\newline$ For $z = \frac{7}{3}$ : $\newline$ $\sqrt{\left(\frac{7}{3}\right)^2 + 8} = 1 - 2\left(\frac{7}{3}\right)$ $\newline$ $\sqrt{\frac{49}{9} + \frac{72}{9}} = 1 - \frac{14}{3}$ $\newline$ $\sqrt{\frac{121}{9}} = 1 - \frac{14}{3}$ $\newline$ $\frac{11}{3} = 1 - \frac{14}{3}$ $\newline$ $\frac{11}{3} = -\frac{11}{3}$ $\newline$ This is not true, so $z = \frac{7}{3}$ is an extraneous solution.For $z = -1$ : $\newline$ $z = \frac{-6}{6}$ $2$ $\newline$ $z = \frac{-6}{6}$ $3$ $\newline$ $z = \frac{-6}{6}$ $4$ $\newline$ $z = \frac{-6}{6}$ $5$ $\newline$ This is true, so $z = -1$ is a valid solution.

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