# How do you complete the square for 2x-3x^2=-8?

Jun 18, 2015

To solve $2 x - 3 {x}^{2} = - 8$ we will collect terms on one side. I prefer a positive in front of ${x}^{2}$, so I'll collect them to make that happen:

$2 x - 3 {x}^{2} = - 8$

$0 = 3 {x}^{2} - 2 x - 8$

Now, of course, this equals that exactly when that equals this, so we can write:

$3 {x}^{2} - 2 x - 8 = 0$

There are a couple of ways to fill in the details of completing the square, but discussing all possibilities is more confusing than helpful, so we'll go through it one way.

$3 {x}^{2} - 2 x - 8 = 0$

On the left, we want a square like ${\left(x - a\right)}^{2} = {x}^{2} - 2 a x + {a}^{2}$. Let's get that $8$ out of our way:

$3 {x}^{2} - 2 x = 8$

We don't want that $3$ out front, so we'll multiply both sides by $\frac{1}{3}$ (Don't forget to distribute on the left.)

$\frac{1}{3} \left(3 {x}^{2} - 2 x\right) = \frac{1}{3} \left(8\right)$

${x}^{2} - \frac{2}{3} x = \frac{8}{3}$

Cookbook: now take $\frac{1}{2}$ of the number in front of $x$, square that and then add that square to both sides. (Why later, in the Note below.)

$\frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}$

${\left(\frac{1}{3}\right)}^{2} = \frac{1}{9}$

Add $\frac{1}{9}$ to both sides:

${x}^{2} - \frac{2}{3} x + \frac{1}{9} = \frac{8}{3} + \frac{1}{9}$

Now we can factor on the left -- remember the $\frac{1}{3}$ we squared? That's what we need now:

${\left(x - \frac{1}{3}\right)}^{2} = \frac{24}{9} + \frac{1}{9}$
Notice that I also got a common denominator on the right so I can do the addition on the right:

${\left(x - \frac{1}{3}\right)}^{2} = \frac{25}{9}$

Now ${a}^{2} = n$ when $a =$ either $\sqrt{n}$ or $- \sqrt{n}$. So

$x - \frac{1}{3} = \pm \sqrt{\frac{25}{9}}$ Simplify the right, to get

$x - \frac{1}{3} = \pm \frac{5}{3}$ Add $\frac{1}{3}$ to both sides:

$x = \frac{1}{3} \pm \frac{5}{3}$

Remember that this means there are two solutions.

One of them is $\frac{1}{3} + \frac{5}{3} = \frac{6}{3} = 2$

and the other is $\frac{1}{3} - \frac{5}{3} = \frac{- 4}{3} = - \frac{4}{3}$

Note

When we got
${x}^{2} - \frac{2}{3} x = \frac{8}{3}$ Why did we do what we did?

We want
${x}^{2} - \frac{2}{3} x + \text{something}$ to be a perfect square like:

${x}^{2} - 2 a x + {a}^{2}$

So the number in front of $x$ is 2 times the thing I need to see the square of. That is $\frac{2}{3} = 2 a$. To fins $a$, take $\frac{1}{2}$ of the number in front of $x$.