How do you factor #x^{2}-8x=2#?

1 Answer
John D.
Oct 13, 2017

#x = 4 +- 3sqrt2#

Explanation:

I'm assuming we need to also solve the equation.

The first step is to move all the terms to one side.

#x^2 - 8x = 2#

#x^2 - 8x - 2 = 0#

Now, to factor this, we need to find two factors of #-2# which add up to #-8#

Through a little bit of trial and error, you will find that there actually aren't any pairs of factors of #-2# that add up to #-8#. We're going to have to solve this another way.

Let's try completing the square. We know that a perfect square has the form:

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

In this case, we have the following:

#2ax = -8x#
#2a = -8#
#a = -4#
#a^2 = 16#

To get #color(blue)16#, we need to add and subtract it from the left side, so it cancels out, like this:

#x^2 - 8x - 2 = 0#

#x^2 - 8x + color(blue)16 - color(blue)16 - 2 = 0#

#(x^2 - 8x + 16) - 18 = 0#

#(x - 4)^2 - 18 = 0#

Now we can continue to solve the equation:

#(x-4)^2 = 18#

#(x-4) = +-sqrt18#

#x = 4 +- sqrt18#

This is our solution! Just one final adjustment to make. Remember that #18 = 2 * 9 = 2 * 3^2#.

#x = 4 +- sqrt(2 * 3^2)#

#x = 4 +- 3sqrt2#

Final Answer

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