# How do you solve the quadratic 3x^2-4x+5-0 by completing the square?

Feb 19, 2015

To start completing the square, you must factor your first coefficient out of your equation (in this case the 3). So pull the 3 out of the first two terms (leave the 3rd one out for now), and this is what you'll get:

$3 \left({x}^{2} - \frac{4}{3} x\right) + 5 = 0$

Then, you need to add something to this equation that makes it a perfect square (which is, after all, what completing the square is).

But what do we need to add? Well, we know that the equation of a perfect square binomial ${\left(a - b\right)}^{2}$ is ${a}^{2} + {b}^{2} - 2 a b$. Therefore, to get the right number, we simply need to divide our 2nd term (without the $x$, as that is the "$a$") by 2 and square it. If we do that, we get:

${\left(\frac{\frac{4}{3}}{2}\right)}^{2} \implies \frac{4}{9}$

Now, we need to add this to our equation. However, keep in mind that because we're adding, we need to add it to the other side as well.

$3 \left({x}^{2} - \frac{4}{3} x + \frac{4}{9}\right) + 5 = \frac{4}{3}$

Now you may have noticed that I added $\frac{4}{3}$ to the other side instead of $\frac{4}{9}$. Why did I do this, you may ask? Well, if you notice carefully, we added the $\frac{4}{9}$ inside the parenthesis. Therefore, we technically also multiplied it by 3. So we must do the same to the other side as well.

Now, you can bring it together as an ${\left(a - b\right)}^{2}$ expression:

$3 {\left(x - \frac{2}{3}\right)}^{2} + 5$ = $\frac{4}{3}$

Now. it's just a matter of simplifying and solving for $x$:

$3 {\left(x - \frac{2}{3}\right)}^{2} = - \frac{11}{3}$
$\left(x - \frac{2}{3}\right) = \pm \sqrt{- \frac{11}{3}}$
$x = \frac{2}{3} \pm i \frac{\sqrt{33}}{3}$