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

Apr 1, 2018

${x}_{1} = - 2 \mathmr{and} {x}_{2} = - 10$

Explanation:

$f \left(x\right) = {x}^{2} + 12 x + 20$

Let $f \left(x\right) = 0$
$0 = {x}^{2} + 12 x + 20$
$0 = {x}^{2} + 2 \cdot 6 \cdot x + {6}^{2} - {6}^{2} + 20$
$0 = {\left(x + 6\right)}^{2} - 36 + 20 = {\left(x + 6\right)}^{2} - 16 | + 16$
$16 = {\left(x + 6\right)}^{2} | \sqrt{}$
$\pm \sqrt{16} = x + 6 | - 6$
$- 6 \pm 4 = {x}_{1 , 2}$
${x}_{1} = - 2 \mathmr{and} {x}_{2} = - 10$

Apr 1, 2018
• $x = - 10$
• $x = - 2$

Explanation:

$f \left(x\right) = {x}^{2} + 12 x + 20$

The roots are the x-intercepts. These occur where $f \left(x\right) = 0$.

$0 = {x}^{2} + 12 x + 20$

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

Factors of $20$: $\left(1 , 20\right) , \left(2 , 10\right) , \left(4 , 5\right)$

If we sum each pair, the one that gives us $12$ is $\left(2 , 10\right)$.

Hence, we factor as:

$0 = \left(x + 10\right) \left(x + 2\right)$

So the roots are found as:

• $x + 10 = 0 \to x = - 10$
• $x + 2 = 0 \to x = - 2$

Hence:

• $x = - 10$
• $x = - 2$
Apr 1, 2018

-2 and - 10

Explanation:

$f \left(x\right) = {x}^{2} + 12 x + 20 = 0$.
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.