Solving Quadratic Equations

Let $x$ and $y$ be the desired real numbers. Using the given information, we can write two equations, one for the sum of the numbers and one for the product of the numbers. $$\begin{array}{rl}\text{Sum: }& x+y=\text{-}13\\ \text{Product: }& xy=42\end{array}$$ We can solve this system of equations using the Substitution Method. First, let's isolate $y$ in the first equation and substitute it into the second equation. The related quadratic equation can be written as $x^2+13x+42=0.$ Our next step is to solve the quadratic equation. We can do this by graphing. Let's begin by finding the axis of symmetry using the formula $x=\text{-}\frac{b}{2a}.$ For our equation, $a=1$ and $b=13.$ Next, we will make a table of values using $x$ values around the axis of symmetry which is $x=\text{-}6.5.$

$x$	$x^2+13x+42$	$y$
$\text{-}11$	$(\text{-}11{)}^2+13(\text{-}11)+42$	$20$
$\text{-}10$	$(\text{-}10{)}^2+13(\text{-}10)+42$	$12$
$\text{-}6.5$	$(\text{-}6.5{)}^2+13(\text{-}6.5)+42$	$\text{-}0.25$
$\text{-}3$	$(\text{-}3{)}^2+13(\text{-}3)+42$	$12$
$\text{-}2$	$(\text{-}2{)}^2+13(\text{-}2)+42$	$20$

Now, we will plot the points and graph the equation that passes through the points.

It seems like the zeros of the equation are $\text{-}7$ and $\text{-}6.$ Let's investigate if they satisfy the original problem. $$\begin{array}{rlrl}\text{Sum: }& \text{-}7+(\text{-}6)=\text{-}13& & ✓\\ \text{Product: }& \text{-}7\cdot (\text{-}6)=42& & ✓\end{array}$$ Since the numbers satisfy both requirements stated in the exercise we have found the numbers we are looking for, and they are $\text{-}7$ and $\text{-}6.$