Question

I was told to put the x value in parenthesis before you square it. can you help?

Answer 1

We have to solve the quadratic equation.

That means that we have to find the roots of it.

The quadratic equation is:

$\begin{gathered} 2x^2-4x-12=0 \\ 2(x^2-2x-6)=0 \\ x^2-2x-6=0 \end{gathered}$

Then, we apply:

$\begin{gathered} x=\frac{-b}{2a}\pm\frac{\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{-(-2)}{2\cdot1}\pm\frac{\sqrt[]{(-2)^2-4\cdot1\cdot(-6)}}{2\cdot1}=\frac{2}{2}\pm\frac{\sqrt[]{4+24}}{2}=1\pm\frac{\sqrt[]{28}}{2}=1\pm\sqrt[]{\frac{28}{4}} \\ x=1\pm\sqrt[]{7} \end{gathered}$$\begin{gathered} x_1=1+\sqrt[]{7}\approx1+2.6=3.6\approx4 \\ x_2=1-\sqrt[]{7}\approx1-2.6=-1.6\approx-1.5 \end{gathered}$

With the rounded values of the roots, Option B is the answer.