which of the following is an extraneous solution of $sqrt{4x + 41}=x + 5$?\n$x=-8$\n$x=-2$\n$x = 2$\n$x = 8$

Explanation:

Step1: Square both sides of the equation

Given $\sqrt{4x + 41}=x + 5$, squaring both sides gives $4x+41=(x + 5)^2$. Expand the right - hand side: $4x+41=x^{2}+10x + 25$. Rearrange to get a quadratic equation: $x^{2}+10x-4x+25 - 41=0$, which simplifies to $x^{2}+6x - 16=0$.

Step2: Factor the quadratic equation

Factor $x^{2}+6x - 16$ as $(x + 8)(x - 2)=0$. Set each factor equal to zero: $x+8=0$ gives $x=-8$ and $x - 2=0$ gives $x = 2$.

Step3: Check the solutions

For $x = 2$: Left - hand side: $\sqrt{4\times2+41}=\sqrt{8 + 41}=\sqrt{49}=7$. Right - hand side: $2 + 5=7$. So $x = 2$ is a valid solution. For $x=-8$: Left - hand side: $\sqrt{4\times(-8)+41}=\sqrt{-32 + 41}=\sqrt{9}=3$. Right - hand side: $-8 + 5=-3$. Since $\sqrt{9}\neq-3$, $x=-8$ is an extraneous solution.

Answer:

$x=-8$