# How do you solve the equation by completing the square x^2-14x+33=0?

Nov 2, 2016

$x = 11$ or $x = 3$.

#### Explanation:

The idea of completing the square, otherwise abbreviated as CTS, is not necessarily complex, but it's a bit involved. First, let's remember that a quadratic equation is of the following format:
$a {x}^{2} + b x + c = 0$, where $a , b$ and $c$ are coefficients of the equation. Coefficient is just a fancy word for numbers attached to the variable (in this case, the variable x). For this equation, we can easily identify our $a , b$ and $c$ numbers! Just eyeball it.

$a = 1$, since ${x}^{2} = 1 \setminus \cdot {x}^{2}$.
$b = - 14$
$c = 33$

Now that we've got that out of the way, let's consider the following algebraic term: $r = {\left(\setminus \frac{b}{2 a}\right)}^{2}$. Our goal is now to calculate $r$. I'll explain what this means a bit later on, but if I do so now, you'll get bogged down.

So, $r = {\left(\setminus \frac{b}{2 a}\right)}^{2} = {\left(\setminus \frac{- 14}{2 \setminus \cdot 1}\right)}^{2} = {\left(- 7\right)}^{2} = 49$

Great. But, now what? Well, we add 49 to both sides of the equation! Analyze how I add it in, though.

$\left({x}^{2} - 14 x + 49\right) + 33 = 49$.

See what I did there? I injected 49 conveniently, right next to -14x. The question you should be asking is, why? Well, this is factorable! In fact, it factors into the following:

${\left(x - 7\right)}^{2} + 33 = 49$
Notice how we used the term $\setminus \frac{b}{2 a} = \setminus \frac{- 14}{2 \setminus \cdot 1} = - 7$ to find the factored form.

Now, let's subtract 33 from both sides.
${\left(x - 7\right)}^{2} = 16$

Now, this is the easy-peezy-lemon-squeazy part. Take the square root of both sides. In doing so, you'll get rid of the exponential 2 on the left side of your equation and on the right side, you'll take the square root of 16. So,

$\setminus \sqrt{{\left(x - 7\right)}^{2}} = \setminus \sqrt{16}$
simplifies to
$x - 7 = \setminus \pm 4$

Which further simplifies to

$x = 7 \setminus \pm 4$.

Your solutions are: ${x}_{1} = 7 + 4 = 11$ or ${x}_{2} = 7 - 4 = 3$. That's about it.

Explanation of the $r = {\left(\setminus \frac{b}{2 a}\right)}^{2}$ term:
This comes from the proof of the quadratic formula. In fact, one of the ways in which we prove the quadratic formula is through the CTS methodology.