Question

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

Answer 1

`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`

Answer 2

-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