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

1 Answer
Nov 2, 2016

Answer:

#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:
#ax^2 + bx + 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 \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 = (\frac{b}{2a})^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 = (\frac{b}{2a})^2 = (\frac{-14}{2\cdot 1})^2 = (-7)^2 = 49#

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

#(x^2 -14x + 49) + 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:

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

Now, let's subtract 33 from both sides.
#(x - 7)^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,

#\sqrt{(x-7)^2} = \sqrt{16}#
simplifies to
#x - 7 = \pm 4#

Which further simplifies to

#x = 7 \pm 4#.

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

Explanation of the #r = (\frac{b}{2a})^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.