Chapter 4.2, Problem 36PPS

To determine

To find: the two real numbers with a sum of $-13$ and a product of $42$ by using quadratic equation.

Answer to Problem 36PPS

  $-7\text{and}-6$ .

Explanation of Solution

Given:

The sum of two real number is $-13$ and its product of $42$ .

Calculation:

Represent the given condition is expressed in to standard form:

Let x be the variable for the first unknown number.

And $-13-x$ for the second unknown number.

Therefore, according to the question,

  $\begin{array}{l}x\left(-13-x\right)=42\{\text{given equation}}\\ -13x-x^2=42\left\{\text{simplify}\right\}\\ -x^2-13x-42=0\left\{\text{equate one side to 0}\right\}\end{array}$ .

Now, graph the related quadratic function to show that x -intercept of the graph. Therefore,

  

Looking at the graph, the parabola crosses the x -axis, therefore, the equation has the solution.

The axis of symmetric is

  $\begin{array}{l}x=\frac{-b}{2a}\left\{\text{equation for symmetry}\right\}\\ x=\frac{-\left(-13\right)}{2\left(-1\right)}\{\text{substitutes these value}}\\ x=\frac{-13}{2}=-6.5\end{array}$

Now write the table for the given function.

$x$	$-8$	$-7$	$-6.5$	$-6$	$-5$
$f\left(x\right)$	$-2$	$0$	$0.25$	$0$	$-2$

Based on the table and the graph, the function has two zero $-7\text{and}-6$ .

Hence, the solution of the given equation is between $-7\text{and}-6$