Question

How do you solve $a^2+11a=-18$?

Answer 1

Step-1: Factor the term using $AC$ method:

Given that $a^2+11a=-18$

Rewrite the term:

$a^2+11a+18=0$

AC Method is a kind of factorization method using variables like $a,b,c.$

Consider the form $x^2+bx+c$:

Find a pair of integers whose product is ${\displaystyle c}$ and whose sum is ${\displaystyle b}$.

In this case of $a^2+11a+18=0$, find the pair of integers whose product is ${\displaystyle 18}$ and the sum is ${\displaystyle 11}$.

$\Rightarrow 2,9[\therefore 2\times 9=18;2+9=11]$

Step-2: Write the integers in factored form:

$$\begin{gathered} a^2+11a+18=0 \\ \Rightarrow a^2+9a+2a+18=0 \\ \Rightarrow a(a+9)+2(a+9)=0 \\ \Rightarrow (a+9)(a+2)=0 \end{gathered}$$

So, the values of $a$ is:

$$\begin{gathered} \Rightarrow a=-2 \\ \Rightarrow a=-9 \end{gathered}$$

Hence, by solving the expression ${\mathit{a}}^{\mathbf{2}}\mathbf{+}\mathbf{11}\mathit{a}\mathbf{=}\mathbf{-}\mathbf{18}$, the values of $\mathit{a}$ are $\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{-}\mathbf{9}\mathbf{.}$