Question

How do you find two integers whose sum is $-8$ and product is $-48$?

Answer 1

Hint: We first assume the two integers. Then we use the given conditions of sum being $-8$ and product being $-48$. We express them in the form of mathematical expression and get a quadratic equation. We solve that to find the integers.

Complete step-by-step solution:
We first assume two integers of a and b.
The sum of those two integers is $-8$ and the product is $-48$.
We express this two information in the form of mathematical expressions.
So, multiplication of a and b gives $-48$ and sum gives $-8$.
Mathematical form gives $a+b=-8$ and $ab=-48$.
We have two equations of two unknowns to solve.
We use the process of substitution.
We can also find the value of one variable $b$ with respect to $a$ based on the equation
$a+b=-8$ where $b=-a-8$. We replace the value of $b$ in the second equation of
$ab=-48$ and get
\[\begin{align}
  & ab=-48 \\
 & \Rightarrow a\left( -a-8 \right)=-48 \\
 & \Rightarrow -{{a}^{2}}-8a=-48 \\
 & \Rightarrow {{a}^{2}}+8a-48=0 \\
\end{align}\]
We get a quadratic equation of a where \[{{a}^{2}}+8a-48=0\].
We solve it using quadratic solving procedure.
We know for a general equation of quadratic $p{{x}^{2}}+bx+c=0$, the value of the roots of x will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4pc}}{2p}$.
In the given equation we have \[{{a}^{2}}+8a-48=0\]. The values of p, b, c are $1,8,-48$ respectively.
We put the values and get a as $a=\dfrac{-8\pm \sqrt{{{8}^{2}}-4\times 1\times \left( -48 \right)}}{2\times 1}=\dfrac{-8\pm \sqrt{256}}{2}=\dfrac{-8\pm 16}{2}=4,-12$.
Putting value of a in $b=-a-8$, we get $b=-a-8=-12,4$.
Therefore, the two integers are $-12,4$.

Note: We can solve the quadratic equation in grouping method.
\[\begin{align}
  & {{a}^{2}}+8a-48=0 \\
 & \Rightarrow {{a}^{2}}+12a-4a-48=0 \\
 & \Rightarrow \left( a+12 \right)\left( a-4 \right)=0 \\
 & \Rightarrow a=-12,4 \\
\end{align}\]