Question

solve by completing the squareX^2 - 4x - 8 = 0

Answer 1

The given expression is

$x^2-4x-8=0$

First, we divide the linear term by 2 and elevate it to the square power.

$(\frac{4}{2})^2=(2)^2=4$

So, we have to add 4 on each side

$\begin{gathered} x^2-4x+4=8+4 \\ x^2-4x+4=12 \end{gathered}$

Then, we factor the trinomial given that is a perfect square trinomial, we take the square root of the first and third terms to express it as the square of a binomial

$(x-2)^2=12$

Now, we solve for x. First, we take the square root on each side

$\begin{gathered} \sqrt[]{(x-2)^2}=\pm\sqrt[]{12} \\ x-2=\pm\sqrt[]{3\cdot4} \\ x=\pm2\sqrt[]{3}+2 \end{gathered}$

Hence, the solutions are

$\begin{gathered} x_1=2\sqrt[]{3}+2 \\ x_2=-2\sqrt[]{3}+2 \end{gathered}$