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

Jun 19, 2015

$2 {x}^{2} - 2 x - 2 = 0$
$2 {x}^{2} - 2 x = 2$
${x}^{2} - x = 1$
${x}^{2} - x + {\left(\frac{1}{2}\right)}^{2} = {\left(\frac{1}{2}\right)}^{2} + 1$
${\left(x - \frac{1}{2}\right)}^{2} = \frac{5}{4}$
$x - \frac{1}{2} = \pm \frac{\sqrt{5}}{2}$
$x = \frac{1 \pm \sqrt{5}}{2}$

But what did we do and why?

In order to have a habit for how we do this, let's begin by moving the constant to the other side of the equation. We'll add $2$ to both sides to get:

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

When completed, the square will be on the left, and it will have the form:

${x}^{2} \pm 2 a x + {a}^{2}$,

so the next thing to do is get a $1$ (which we won't write) in front of the ${x}^{2}$. (In fancy terms, we're going to make the coefficient of ${x}^{2}$ equal to $1$.)

Multiply both sides of the equation by $\frac{1}{2}$ (Remember to distribute the multiplication.)

$\frac{1}{2} \left(2 {x}^{2} - 2 x\right) = \frac{1}{2} \left(2\right)$ now simplify:

${x}^{2} - x = 1$

Now that we have just ${x}^{2}$, we can see that the coefficient of $x$ is negative, that tells us that the completed square will look like:

${x}^{2} - 2 a x + {a}^{2}$ which we will be able to factor: ${\left(x - a\right)}^{2}$

We have: ${x}^{2} - x = 1$,
Which we can think of as: ${x}^{2} - 1 x = 1$,

so we must have:

$2 a = 1$.

And that makes $a = \frac{1}{2}$. Square that and add the result to both sides:
${\left(\frac{1}{2}\right)}^{2} = {1}^{2} / {2}^{2} = \frac{1}{4}$

${x}^{2} - x + \frac{1}{4} = \frac{1}{4} + 1$,

Now factor on the left (we already know how to factor that! See above.) And add on the right.

${\left(x - \frac{1}{2}\right)}^{2} = \frac{1}{4} + \frac{4}{4} = \frac{5}{4}$

${\left(x - \frac{1}{2}\right)}^{2} = \frac{5}{4}$

Now use the fact that ${n}^{2} = g$ if and only of $n = \sqrt{g} \mathmr{and} - \sqrt{g}$
(The square of a number equals a given number if and only if the number is either the positive or negative square root of the given.)

$x - \frac{1}{2} = \pm \sqrt{\frac{5}{4}} = \pm \frac{\sqrt{5}}{\sqrt{4}} = \pm \frac{\sqrt{5}}{2}$

$x - \frac{1}{2} = \pm \frac{\sqrt{5}}{2}$ Add $\frac{1}{2}$ to both sides:

$x = \frac{1}{2} \pm \frac{\sqrt{5}}{2}$ which we often prefer to write as a single fraction:

$x = \frac{1 \pm \sqrt{5}}{2}$

Note: never let yourself think this is some kind pf weird positive and negative number. It is just a convenient way of writing the two solutions:
$x = \frac{1 + \sqrt{5}}{2}$ and $x = \frac{1 - \sqrt{5}}{2}$.