Question

Which of the following is an extraneous solution of 4x+41 - X+5?O x= -8O x = -2O x= 2O x = 8

Answer 1

Given the equation:

$\sqrt[]{4x+41}=x+5$

Square the both sides

$\begin{gathered} 4x+41=(x+5)^2 \\ 4x+41=x^2+10x+25 \\ x^2+10x+25-4x-41=0 \\ x^2+6x-16=0 \\ (x+8)(x-2)=0 \\ x+8=0\rightarrow x=-8 \\ x-2=0\rightarrow x=2 \end{gathered}$

now, we will check x = 2

$\begin{gathered} \sqrt[]{4x+41}=\sqrt[]{4\cdot2+41}=\sqrt[]{49}=7 \\ x+5=2+5=7 \end{gathered}$

So, x = 2 is a solution to the equation

We will check x = -8

$\begin{gathered} \sqrt[]{4x+41}=\sqrt[]{4\cdot-8+41}=\sqrt[]{9}=3 \\ x+5=-8+5=-3 \end{gathered}$

so, x = -8 is an extraneous solution

So, the answer is x = -8