# How do you solve using the completing the square method 2x^2-4x-14=0?

Apr 28, 2018

The two solutions are $x = \pm 2 \sqrt{2} + 1$.

#### Explanation:

First, divide the whole equation by two:

$2 {x}^{2} - 4 x - 14 = 0$

${x}^{2} - 2 x - 7 = 0$

Then, move the constant to the other side:

${x}^{2} - 2 x = 7$

Next, identify the square binomial on the left side. We know that ${\left(x - 1\right)}^{2}$ expands to ${x}^{2} - 2 x + 1$, so if we add $1$ to both sides, we can use this backwards:

${x}^{2} - 2 x \textcolor{red}{+} \textcolor{red}{1} = 7 \textcolor{red}{+} \textcolor{red}{1}$

${\left(x - 1\right)}^{2} = 8$

Now, square root both sides:

$x - 1 = \pm \sqrt{8}$

$x - 1 = \pm 2 \sqrt{2}$

$x = \pm 2 \sqrt{2} + 1$

Those are the solutions. Hope this helped!

Apr 28, 2018

$x = 1 \pm 2 \sqrt{2}$

#### Explanation:

Factor out 2:

$2 \left[{x}^{2} - 2 x\right] - 14$ $\Leftarrow$ notice we didn't factor the constant.

Complete the square of the brackets:

$2 \left[{\left(x - 1\right)}^{2} - 1\right] - 14$

Multiply the brackets:

$2 {\left(x - 1\right)}^{2} - 2 - 14$

Simplify:

$2 {\left(x - 1\right)}^{2} - 16$

We can always check this:

${\left(x - 1\right)}^{2} = {x}^{2} - 2 x + 1$

$2 \left({x}^{2} - 2 x + 1\right) = 2 {x}^{2} - 4 x + 2 - 16 \to 2 {x}^{2} - 4 x - 14$

Solving:

$2 {\left(x - 1\right)}^{2} - 16 = 0$

$2 {\left(x - 1\right)}^{2} = 16$

${\left(x - 1\right)}^{2} = 8$

$x - 1 = \pm \sqrt{8}$

$x = 1 \pm \sqrt{8}$

$\to x = 1 \pm 2 \sqrt{2}$