Zhang Li solves the equation below by first squaring both sides of the equation. -4=sqrt(5z+7) What extraneous solution does Zhang Li obtain? z=

Square both sides: Square both sides of the equation to eliminate the square root. -4 = sqrt(5z + 7) (-4)^2 = (sqrt(5z + 7))^2 16 = 5z + 7Subtract 7 to isolate z: Subtract 7 from both sides to isolate the term with z. 16 - 7 = 5z + 7 - 7 9 = 5zDivide both sides by 5: Divide both sides by 5 to solve for z. 9 / 5 = 5z / 5 z = 9 / 5Check the solution: Check the solution by substituting it back into the original equation. -4 = sqrt(5(9/5) + 7) -4 = sqrt(9 + 7) -4 = sqrt(16) -4 = 4 or -4 = -4 Here we see that the original equation has a negative number on one side and a square root on the other side, which cannot be negative. Therefore, the solution z = 9/5 is extraneous because it leads to a false statement when substituted back into the original equation.

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Zhang Li solves the equation below by first squaring both sides of the equation.

-4=sqrt(5z+7)
What extraneous solution does Zhang Li obtain?

z=

Zhang Li solves the equation below by first squaring both sides of the equation. $\newline$ $-4=\sqrt{5 z+7}$ $\newline$ What extraneous solution does Zhang Li obtain? $\newline$ $z=$

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

\newline

-4=\sqrt{5 z+7}

\newline

What extraneous solution does Zhang Li obtain?

\newline

z=

Square both sides: Square both sides of the equation to eliminate the square root. $\newline$ $-4 = \sqrt{5z + 7}$ $\newline$ $(-4)^2 = (\sqrt{5z + 7})^2$ $\newline$ $16 = 5z + 7$
Subtract $7$ to isolate $z$ : Subtract $7$ from both sides to isolate the term with $z$ . $16 - 7 = 5z + 7 - 7$ $9 = 5z$
Divide both sides by $5$ : Divide both sides by $5$ to solve for $z$ . $\frac{9}{5} = \frac{5z}{5}$ $z = \frac{9}{5}$
Check the solution: Check the solution by substituting it back into the original equation. $\newline$ $-4 = \sqrt{5(\frac{9}{5}) + 7}$ $\newline$ $-4 = \sqrt{9 + 7}$ $\newline$ $-4 = \sqrt{16}$ $\newline$ $-4 = 4$ or $-4 = -4$ $\newline$ Here we see that the original equation has a negative number on one side and a square root on the other side, which cannot be negative. Therefore, the solution $z = \frac{9}{5}$ is extraneous because it leads to a false statement when substituted back into the original equation.