How do you solve #x^2 + 12x + 20 = 0# by completing the square?

2 Answers
Mar 3, 2018

Answer:

#x=-10, x=-2#

Explanation:

Our Quadratic is in the form #ax^2+bx+c=0#. When we complete the square, we take half of the #b# value, square it, and add it to both sides.

First, let's subtract #20# from both sides. We get:

#x^2+12x =-20#

Half of #12# is #6#, and when we square #6#, we get #36#. Let's add this to both sides. We get:

#x^2+12x+36=-20+36#

Simplifying, we get:

#x^2+12x+36=16#

Now, we factor the left side of the equation. We think of two numbers that add up to #12# and have a product of #36#. #6# and #6# are our two numbers.

We can now factor this as:

#color(blue)((x+6)(x+6))=16#

Notice, what I have in blue is the same as #x^2+12x+36#. We can rewrite this as:

#(x+6)^2=16#

Taking the square root of both sides, we get:

#x+6=-4# and #x+6=4#

Subtracting #6# from both sides of the equations, we get:

#x=-10, x=-2#

Mar 3, 2018

Answer:

#x=-2# and #x=-10#

Explanation:

By completing the square, we always half the coefficient of #x#, putting it in brackets and squaring it, as a quadratic is always in the form #ax^2+bx+c#. Always remember to add the constant of #(+20)# on the end:

#therefore# #x^2+12x -> (x+6)^2#

Also by completing the square, we also take away the squared number of half of the previous coefficient (the number in the brackets).

#(x+6)^2-36+20#

Simplifying terms:

#-36+20=-16#

Plugging this back in, removing the #-36# and #20#:

#(x+6)^2-16#

This is in the completed square form. You can always check this by expanding out:

#(x+6)(x+6) -> x^2+6x+6x+36-16 -> x^2+12x+20#

#therefore# This complete the square form is correct.

Solving; so far we know:

#(x+6)^2-16=0#

So we add #16# to solve this, as we cannot solve when equal to 0.

#(x+6)^2=16#

As we do not want squares brackets, the opposite of squaring is square rooting, and this cancels out the squared brackets:

#(x+6)^2=16 -> x+6=pmsqrt16#

Always remember the #pm# when square rooting...

As we want #x# on its own, we have to #-6# to get rid of it, transferring the #-6# to the other side of the equation.

#x+6=pmsqrt16 -> x=-6pmsqrt16#

Always remember there are mostly two solutions, as there is a plus and minus on the answer.

#therefore# our answers are...

#x=-6pmsqrt16# and #x=-6pmsqrt16#

But using our squared numbers:

#1^2=1xx1=1#
#2^2=2xx2=4#
#3^2=3xx3=9#
#4^2=4xx4=16# #sqrt16# can be simplified further as it is a squared number...

#x=-6pmsqrt16 -> x=-6pm4#

#therefore# the answers are:

#x=-6+4# and #x=-6-4# since most of the time we have two solutions.

These simplify to:

#x=-2# and #x=-10#