How do you solve #x^2 = 4x + 12#?

2 Answers
Jul 12, 2016

Answer:

#x = 6# and #x = -2#

Explanation:

By moving #4x# to the left of our equation, we can then complete the square in the following way:

#x^2 = 4x + 12#

#x^2 -4x = 12#

We take the coefficient on the #x#-term, namely #-4#, divide it by #2#, and square the result, giving us

#(-4/2)^(2) = (-2)^2 = 4#

Since our goal is to rewrite our equation in the form of

#x^2 -4x + ? = 12 + ?#

We replace our #?# marks with the result of #4# we just calculated, giving us

#x^2-4x+4 = 16#

Now, we are looking for two numbers whose product gives #4# and when added together gives us #-4#.

We can see that #-2 * -2= 4# and #-2 + -2 = -4#, so our factors are the numbers #-2# and #-2#, thus we can rewrite our equation in the form of

#(x-2)(x-2) = 16#, or simply #(x-2)^(2) = 16#

Taking the square root of both sides yields

#x-2 = ± sqrt(16)#

Adding #2# to both sides then gives us

#x = ± 4 + 2#

So our solutions are

#x = 6# and #x = -2#

Jul 12, 2016

Answer:

-2 and 6

Explanation:

#x^2 - 4x - 12 = 0#
Find 2 numbers (real roots), that have opposite signs, knowing the sum (-b = 4) and the product (c = -12). They make the factor pair
(-2, 6) --> sum (4) and product (-12).
2 real roots: -2, and 6