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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.\newlinez2+8=1−2z \sqrt{z^{2}+8}=1-2 z \newlineWhat extraneous solution does Avery obtain?\newlinez= z=

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Q. Avery solves the equation below by first squaring both sides of the equation.\newlinez2+8=1−2z \sqrt{z^{2}+8}=1-2 z \newlineWhat extraneous solution does Avery obtain?\newlinez= z=
  1. 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?＂
  2. Squaring both sides: Avery starts with the equation z2+8=1−2z\sqrt{z^2 + 8} = 1 - 2z. To eliminate the square root, Avery squares both sides of the equation.\newline(\sqrt{z^\(2\) + \(8\)})^\(2 = (11 - 22z)^22
  3. Expanding the equation: After squaring both sides, the equation becomes: z2+8=(1−2z)2z^2 + 8 = (1 - 2z)^2
  4. Rearranging the equation: Next, Avery expands the right side of the equation:\newlinez2+8=12−2(1)(2z)+(2z)2z^2 + 8 = 1^2 - 2(1)(2z) + (2z)^2\newlinez2+8=1−4z+4z2z^2 + 8 = 1 - 4z + 4z^2
  5. Combining like terms: Now, Avery rearranges the equation to set it to zero:\newline0=4z2−4z+1−z2−80 = 4z^2 - 4z + 1 - z^2 - 8
  6. Using the quadratic formula: Avery combines like terms: 0=3z2−4z−70 = 3z^2 - 4z - 7
  7. Simplifying the square root: Avery uses the quadratic formula to solve for zz:
    z=−(−4)±(−4)2−4(3)(−7)2⋅3z = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(-7)}}{2 \cdot 3}
    z=4±16+846z = \frac{4 \pm \sqrt{16 + 84}}{6}
    z=4±1006z = \frac{4 \pm \sqrt{100}}{6}
  8. Finding the solutions: Avery simplifies the square root and continues solving:\newlinez=4±106z = \frac{4 \pm 10}{6}\newlineThis gives two solutions: z=(4+10)6z = \frac{(4 + 10)}{6} and z=(4−10)6z = \frac{(4 - 10)}{6}
  9. Checking the solutions: Avery finds the two solutions:\newlinez = 146\frac{14}{6} and z = −66\frac{-6}{6}\newlinez = 73\frac{7}{3} and z = −1-1
  10. Checking the solutions: Avery finds the two solutions:\newlinez=146z = \frac{14}{6} and z=−66z = \frac{-6}{6}\newlinez=73z = \frac{7}{3} and z=−1z = -1Avery checks both solutions by substituting them back into the original equation:\newlineFor z=73z = \frac{7}{3}:\newline(73)2+8=1−2(73)\sqrt{\left(\frac{7}{3}\right)^2 + 8} = 1 - 2\left(\frac{7}{3}\right)\newline499+729=1−143\sqrt{\frac{49}{9} + \frac{72}{9}} = 1 - \frac{14}{3}\newline1219=1−143\sqrt{\frac{121}{9}} = 1 - \frac{14}{3}\newline113=1−143\frac{11}{3} = 1 - \frac{14}{3}\newline113=−113\frac{11}{3} = -\frac{11}{3}\newlineThis is not true, so z=73z = \frac{7}{3} is an extraneous solution.
  11. Checking the solutions: Avery finds the two solutions:\newlinez=146z = \frac{14}{6} and z=−66z = \frac{-6}{6}\newlinez=73z = \frac{7}{3} and z=−1z = -1Avery checks both solutions by substituting them back into the original equation:\newlineFor z=73z = \frac{7}{3}:\newline(73)2+8=1−2(73)\sqrt{\left(\frac{7}{3}\right)^2 + 8} = 1 - 2\left(\frac{7}{3}\right)\newline499+729=1−143\sqrt{\frac{49}{9} + \frac{72}{9}} = 1 - \frac{14}{3}\newline1219=1−143\sqrt{\frac{121}{9}} = 1 - \frac{14}{3}\newline113=1−143\frac{11}{3} = 1 - \frac{14}{3}\newline113=−113\frac{11}{3} = -\frac{11}{3}\newlineThis is not true, so z=73z = \frac{7}{3} is an extraneous solution.For z=−1z = -1:\newlinez=−66z = \frac{-6}{6}22\newlinez=−66z = \frac{-6}{6}33\newlinez=−66z = \frac{-6}{6}44\newlinez=−66z = \frac{-6}{6}55\newlineThis is true, so z=−1z = -1 is a valid solution.

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