Question

How do you find the roots for `f(x) = x^2 + 12x + 20`?

Answer 1

`x_1=-2 or x_2=-10`

Explanation:

`f(x) = x^2 + 12x + 20`

Let `f(x)=0`
`0 = x^2 + 12x + 20`
`0 = x^2 + 2*6*x + 6^2-6^2+20`
`0 = (x+6)^2-36+20= (x+6)^2-16|+16`
`16 = (x+6)^2|sqrt()`
`+-sqrt(16)=x+6|-6`
`-6+-4=x_(1,2)`
`x_1=-2 or x_2=-10`

Answer 2

• `x = -10`
• `x = -2`

Explanation:

`f(x) = x^2 +12x +20`

The roots are the x-intercepts. These occur where `f(x) = 0`.

`0 = x^2 + 12x + 20`

Now we can factor the right side. We need to find factors of `20` that when summed give us `12`.

Factors of `20`: `(1,20), (2,10),(4,5)`

If we sum each pair, the one that gives us `12` is `(2,10)`.

Hence, we factor as:

`0 = (x+10)(x+2)`

So the roots are found as:

• `x+10=0 -> x = -10`
• `x+2=0 -> x = -2`

Hence:

• `x = -10`
• `x = -2`

Answer 3

-2 and - 10

Explanation:

`f(x) = x^2 + 12x + 20 = 0`.
Find 2 real roots, that are both negative (ac > 0; ab > 0), knowing their sum (- b = - 12) and their product (c = 20).
They are: -2 and - 10.

Note . When a = 1, we don't have to do factoring by grouping and solving the 2 binomials.