Chapter 12.6, Problem 3P

To determine

To find: the sum of a number whose square is equal to $56$ .

Answer to Problem 3P

The numbers are $7\&-8$ .

Explanation of Solution

Given information:

The sum of a number whose square is equal to $56$ .

Formula:

The quadratic equation formula $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ where $a,b,c$ is any constant.

Calculation:

Consider the number be $x$ .

Then as mentioned above that the sum of a number whose square is equal to $56$ .

Therefore $x+x^2=56$ .

Now $x^2+x-56=0$ is in the form of $a{x}^2+bx+c=0$ where $a=1,b=1,c=-56$ .

Recall that quadratic equation formula $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ where $a,b,c$ is any constant.

Therefore

  $\begin{array}{l}{x}^2+x-56=0\Rightarrow x=\frac{-1\pm \sqrt{1^2+4*1*56}}{2}\Rightarrow x=\frac{-1\pm \sqrt{1+224}}{2}\Rightarrow x=\frac{-1\pm \sqrt{225}}{2}\\ \Rightarrow x=\frac{-1\pm 15}{2}\Rightarrow x=\frac{-1-15}{2},\frac{-1+15}{2}\Rightarrow x=-8,7.\end{array}$

The sum of numbers whose square is $56$ are $7\&-8$ .