How do you factor and solve # x^2-12x=-6 #?

1 Answer
Ryuu
Apr 29, 2017

Bring over the #-6# and factor as you would a simple trinomial.

In this case, that doesn't work (it's "unfactorable"), and you have to use the quadratic formula to solve.

As a result, you'll get #x~=11.477# and #x~=0.523#.

Explanation:

Factoring

In order to factor, let's first bring the #-6# over, and equate the equation to #0#.

#x^2 - 12x = - 6#

#x^2 - 12x + 6= 0#

Now, let's factor it as you would a simple trinomial. Meaning, "what two numbers multiplied equals #ac# and added equals #b#?"

There are no numbers that fits this requirement (^). Therefore, it is "unfactorable". Due to this, we have to use the second way to factor: quadratic equation.

This ultimately brings us to...

Solving.

#x=\frac{-b\pm\sqrt{b^2-4ac\ }}{2a}#

Now let's sub in the values.

#x=\frac{-(-12)\pm\sqrt{(-12)^2-4(1)(6)\ }}{2(1)}#

#x=\frac{12\pm\sqrt{120\ }}{2}#

At this point, we can solve for #x#. We will get two answers because of the #+-# sign.

#x=\frac{12\pm\sqrt{120\ }}{2}#

  1. #x=\frac{12\+\sqrt{120}}{2}#

#x~=\frac{22.954}{2}#

#x~=11.477#

2.
#x=\frac{12\-\sqrt{120}}{2}#

#x~=\frac{1.046}{2}#

#x~=0.523#

We can graph the equation to check our work.

graph{x^2 - 12x + 6 [-2.8, 22.51, -6.33, 6.33]}

As you can see, the zeros match up.

Hope this helps :)

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