# Let h(x)=12x+x^2, how do you find a such that h(a)=-27?

Nov 14, 2016

$a = - 9 \mathmr{and} a = - 3$

#### Explanation:

$h \left(a\right) = 12 a + {a}^{2} = - 27 \mathmr{and} {a}^{2} + 12 a + 27 = 0 \mathmr{and} \left(a + 9\right) \left(a + 3\right) = 0$. Either $a + 9 = 0 \mathmr{and} a + 3 = 0 \therefore a = - 9 \mathmr{and} a = - 3$ [Ans]

Nov 14, 2016

$a = - 3 , a = - 9$

#### Explanation:

Express h(x) in terms of a.

That is $h \left(\textcolor{red}{a}\right) = 12 \textcolor{red}{a} + {\left(\textcolor{red}{a}\right)}^{2} = 12 a + {a}^{2}$

$h \left(a\right) = - 27 \text{ and } h \left(a\right) = 12 a + {a}^{2}$

$\text{solve " 12a+a^2=-27" to find a}$

since this is a quadratic function, equate to zero.

$\Rightarrow {a}^{2} + 12 a + 27 = 0$

using the a-c method, we require the product of factors of 27 that also sum to + 12. These are +3 and +9.

$\Rightarrow \left(a + 3\right) \left(a + 9\right) = 0$

solve : $a + 3 = 0 \Rightarrow a = - 3$

solve : $a + 9 = 0 \Rightarrow a = - 9$

Check :

$a = - 3 \Rightarrow 12 \times \left(- 3\right) + {\left(- 3\right)}^{2} = - 36 + 9 = - 27 \textcolor{w h i t e}{x}$

$a = - 9 \Rightarrow 12 \times \left(- 9\right) + {\left(- 9\right)}^{2} = - 108 + 81 = - 27$

$\Rightarrow a = - 3 , a = - 9 \text{ are the solutions}$