How do you solve using the completing the square method #x^2 - 5x = 9#?

1 Answer
Jan 20, 2017

Answer:

Take the coefficient of the #x# term (#-5#) and:
1. Divide it by 2 (to get #-5/2#).
2. Square this (to get #25/4#).
3. Add this final value to both sides.

Explanation:

Completing the square means seeking a constant term #n# to add to #x^2-5x#, so that #x^2-5x+n# is a perfect square.

First, let's look at what happens when we FOIL a perfect square binomial of the form #(x+a)^2#:

#(x+a)(x+a)=x^2+2ax+a^2#

A perfect square will always have a distributed form like this.

What we notice is that if we take the coefficient of the #x# term (#2a#), cut it in half, and then square it, we get #a^2#, the constant term. Thus, if given #x^2+2ax=b#, we would complete the square by adding #a^2# (that is, the square of half of #2a#) to both sides, giving

#x^2+2ax+a^2 = b + a^2#

so that the trinomial on the left will be guaranteed to be a perfect square—the square of #(x+a)#.

For this particular problem, we are given #x^2-5x=9#. So, #-5# is like our "#2a#". And if

#-5=2a#,

then

#a=-5/2#,

and

#a^2=25/4#.

Thus, #x^2-5x+25/4# is the completed square we seek, meaning we need to add #25/4# to both sides:

#x^2-5x+25/4=9+25/4#

Okay, so if this #x^2-5x+25/4# is a perfect square, what is its factored form? (Or, what is its square root?)

That's easy—remember that the factored form of #x^2+color(red)(2a)x+a^2# is #(x+color(blue)a)^2#. The #color(blue)a# in the factor is half of the #color(red)(2a)# in the trinomial. So the factored form of #x^2-5x+25/4# will be #(x-5/2)^2#, because #-5/2# is half of #-5#.

We simplify both sides now to get

#(x-5/2)^2=61/4#

Now our LHS is a perfect square, so we can solve for #x# by taking the square root of both sides:

#x-5/2=+-sqrt61 /2#

and then adding #5/2# to both sides:

#x=5/2+-sqrt61 / 2" "=" "(5+-sqrt61)/2#.

Note:

If the coefficient on the #x^2# term is something other than #1#, you'll want to either factor it out or divide everything by it first, so that this method will work.