two possible solutions of $sqrt{11 - 2x}=sqrt{x^{2}+4x + 4}$ are -7 and 1. which statement is true?\nonly…

two possible solutions of $sqrt{11 - 2x}=sqrt{x^{2}+4x + 4}$ are -7 and 1. which statement is true?\nonly $x=-7$ is an extraneous solution.\nonly $x = 1$ is an extraneous solution.\nboth solutions are extraneous.\nneither solution is extraneous.

two possible solutions of $sqrt{11 - 2x}=sqrt{x^{2}+4x + 4}$ are -7 and 1. which statement is true?\nonly $x=-7$ is an extraneous solution.\nonly $x = 1$ is an extraneous solution.\nboth solutions are extraneous.\nneither solution is extraneous.

Answer

Explanation:

Step1: Simplify right - hand side

Since $x^{2}+4x + 4=(x + 2)^{2}$, the equation $\sqrt{11-2x}=\sqrt{x^{2}+4x + 4}$ becomes $\sqrt{11-2x}=\sqrt{(x + 2)^{2}}=\vert x + 2\vert$.

Step2: Check $x=-7$

Substitute $x=-7$ into the original equation. Left - hand side: $\sqrt{11-2\times(-7)}=\sqrt{11 + 14}=\sqrt{25}=5$. Right - hand side: $\sqrt{(-7)^{2}+4\times(-7)+4}=\sqrt{49-28 + 4}=\sqrt{25}=5$. So $x=-7$ is a valid solution.

Step3: Check $x = 1$

Substitute $x = 1$ into the original equation. Left - hand side: $\sqrt{11-2\times1}=\sqrt{9}=3$. Right - hand side: $\sqrt{1^{2}+4\times1+4}=\sqrt{1 + 4+4}=\sqrt{9}=3$. So $x = 1$ is a valid solution.

Answer:

Neither solution is extraneous.