Question

Find all the values of x satisfying the given conditions

Answer 1

The solution set is given by solving the equation:

$x+\sqrt{x+4}=8$

Subtract x from both sides of the equation:

$\begin{gathered} \sqrt{x+4}=8-x \\ \text{ Square both sides of the equation:} \\ x+4=(8-x)^2 \\ x+4=x^2-16x+64 \\ \text{ Therefore,} \\ x^2-17x+60=0 \end{gathered}$

To factorize the quadratic equation, Find two numbers such that their product is 60 and their sum is -17.

The two numbers are 5 and 12.

Hence,

$\begin{gathered} x^2-17x+60=0 \\ x^2-5x-12x+60=0 \end{gathered}$

Factorizing the equation we have:

$x(x-5)-12(x-5)$

Hence,

$\begin{gathered} (x-12)(x-5)=0 \\ \text{ Thus} \\ x=2,5 \end{gathered}$

Check if x = 2 the solutions are extraneous:

$2+\sqrt{2+4}=2+\sqrt{6}$

Hence x=2 is an extraneous solution

Check if x = 5 the solutions are extraneous:

$5+\sqrt{3+4}=5+\sqrt{9}=5+3=8$

Hence x=5 is a solution

Therefore, the solution set is { 5 }