Chapter 4.2, Problem 33PPS

To determine

To find: the two real numbers with a sum of $-15$ and a product of $-54$ by using quadratic equation.

Answer to Problem 33PPS

  $x=-18,x=3$ .

Explanation of Solution

Given:

The sum of two real number is $-15$ and its product of $-54$ .

Calculation:

Represent the given condition is expressed in to standard form:

Let x be the variable for the first unknown number.

And $-15-x$ for the second unknown number.

Therefore, according to the question,

  $\begin{array}{l}x\left(-15-x\right)=-54\{\text{given equation}}\\ -15x-x^2=-54\left\{\text{simplify}\right\}\\ -15x-x^2+54=0\left\{\text{equate one side to 0}\right\}\\ \therefore x^2+15x-54=0\{\text{multiply both side by }-\text{1}\}\end{array}$ .

Now, solve the quadratic equation. Therefore,

  $\begin{array}{l}{x}^2+15x-54=0\left\{\text{given function}\right\}\\ x^2-3x+18x-54=0\{arrangethe\text{factor}}\\ x\left(x-3\right)+18\left(x-3\right)=0\{arrangethe\text{factor}}\\ \left(x-3\right)\left(x+18\right)=0\{\text{factor}}\end{array}$

Therefore, either $\left(x-3\right)=0$ or $\left(x+18\right)=0$

When $\left(x-3\right)=0$ , then $x=3$ .

When $\left(x+18\right)=0$ , then $x=-18$ .

Thus, the zeros of the function are $x=-18,x=3$ .

Hence, the solution of the given equation is $x=-18,x=3$ .