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

Jul 12, 2016

Answer:

$x = 6$ and $x = - 2$

Explanation:

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

${x}^{2} = 4 x + 12$

${x}^{2} - 4 x = 12$

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

${\left(- \frac{4}{2}\right)}^{2} = {\left(- 2\right)}^{2} = 4$

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

${x}^{2} - 4 x + 4 = 16$

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

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

$\left(x - 2\right) \left(x - 2\right) = 16$, or simply ${\left(x - 2\right)}^{2} = 16$

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

-2 and 6

Explanation:

${x}^{2} - 4 x - 12 = 0$
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