How do you solve $2 x^{2} - 2 x - 2 = 0$ $2x^2 -2x- 2= 0$ using completing the square?

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How do you solve $2 x^{2} - 2 x - 2 = 0$ $2x^2 -2x- 2= 0$ using completing the square?

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Answer 1 · 2015-06-19T13:24:14

Quick answer:
$2 x^{2} - 2 x - 2 = 0$ $2x^2-2x-2=0$
$2 x^{2} - 2 x = 2$ $2x^2-2x = 2$
$x^{2} - x = 1$ $x^2 - x = 1$
$x^{2} - x + {(\frac{1}{2})}^{2} = {(\frac{1}{2})}^{2} + 1$ $x^2 - x + (1/2)^2= (1/2)^2+ 1$
${(x - \frac{1}{2})}^{2} = \frac{5}{4}$ $(x-1/2)^2 = 5/4$
$x - \frac{1}{2} = \pm \frac{\sqrt{5}}{2}$ $x-1/2 = +-sqrt5/2$
$x = \frac{1 \pm \sqrt{5}}{2}$ $x=(1+-sqrt5)/2$

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 $2$ $2$ to both sides to get:

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

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

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

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

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

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

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

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

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

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

so we must have:

$2 a = 1$ $2a = 1$ .

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

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

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

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

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

Now use the fact that $n^{2} = g$ $n^2 = g$ if and only of $n = \sqrt{g} or - \sqrt{g}$ $n = sqrtg or -sqrtg$
(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.)

$x - \frac{1}{2} = \pm \sqrt{\frac{5}{4}} = \pm \frac{\sqrt{5}}{\sqrt{4}} = \pm \frac{\sqrt{5}}{2}$ $x-1/2 = +-sqrt(5/4) = +-sqrt5/sqrt4 = +-sqrt5/2$

$x - \frac{1}{2} = \pm \frac{\sqrt{5}}{2}$ $x-1/2 = +-sqrt5/2$ Add $\frac{1}{2}$ $1/2$ to both sides:

$x = \frac{1}{2} \pm \frac{\sqrt{5}}{2}$ $x=1/2 +- sqrt5/2$ which we often prefer to write as a single fraction:

$x = \frac{1 \pm \sqrt{5}}{2}$ $x=(1+-sqrt5)/2$

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:
$x = \frac{1 + \sqrt{5}}{2}$ $x=(1+sqrt5)/2$ and $x = \frac{1 - \sqrt{5}}{2}$ $x=(1-sqrt5)/2$ .

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