Q37. - Algebra 1

Question

Find two numbers that have a sum of $-2$ and a product of $-48$. 

Step-by-Step Solution

Answer

The two numbers are $\mathbf{-}\mathbf{8}$ and 6. 

Step 1. Understand the concept used .

If $a{x}^2+bx+c=0$, with $a\ne 0$, then $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$. Solving using factoring can be done by splitting the middle term of the equation and then making groups and equating them to zero.

Step 2. Substitute the values .

Let $x$ and $y$ be two numbers. Then the system of equations formed as per the information given is:

  $$\begin{gathered} x+y=-2 \dots \left(1\right) \\ x\times y=-48 \dots \left(2\right) \end{gathered}$$

Write equation (1) in terms of $x$.

  $$\begin{gathered} x+y=-2 \\ x+y-y=-2-y \\ x=-2-y \end{gathered}$$

Now substitute the value of $x$ into equation (2).

$$\begin{gathered} x\times y=-48 \\ (-2-y)·y=-48 \\ -2y-y^2=-48\left[\text{\hspace{0.17em}Distribute\hspace{0.17em}\hspace{0.17em}}\right] \end{gathered}$$

Step 3. Solve the quadratic equation.

Simplify the quadratic equation for $y$. by completing the square.

 $-2y+2y-y^2+y^2=-48+2y+y^2\left[\text{\hspace{0.17em}\hspace{0.17em}make\hspace{0.17em}\hspace{0.17em}one\hspace{0.17em}\hspace{0.17em}side\hspace{0.17em}\hspace{0.17em}equal\hspace{0.17em}\hspace{0.17em}to\hspace{0.17em}\hspace{0.17em}zero\hspace{0.17em}\hspace{0.17em}}\right]$

                                $$\begin{gathered} 0=y^2+2y-48 \\ 0=(y-6)(y+8)\left[\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Factor\hspace{0.17em}\hspace{0.17em}}\right] \end{gathered}$$ 

Use the zero product rule to further find the values of $y$.

 $y-6=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}y+8=0$

       $$\begin{gathered} y=6\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}y=-8 \\ x=-2-y \\ x=-2-6\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}x=-2-(-8) \\ x=-8\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}x=6 \end{gathered}$$

The set of numbers is: $(-8,6)$ or $(6,-8)$.