Chapter 12.6, Problem 4P

To determine

To find: the difference of a number and its square is $182$ .

Answer to Problem 4P

The number are $14\&-13$ .

Explanation of Solution

Given information:

The difference of a number and its square is $182$ .

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 difference of a number and its square is $182$ .

Therefore $x^2-x=182$ .

Now $x^2-x=182\Rightarrow x^2-x-182=0$ is in the form of $a{x}^2+bx+c=0$ where $a=1,b=-1,c=-182$ .

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

Therefore

  $\begin{array}{l}{x}^2-x-182=0\Rightarrow x=\frac{-\left(-1\right)\pm \sqrt{{\left(-1\right)}^2+4*1*182}}{2}\Rightarrow x=\frac{1\pm \sqrt{1+728}}{2}\Rightarrow x=\frac{1\pm \sqrt{729}}{2}\\ \Rightarrow x=\frac{1\pm 27}{2}\Rightarrow x=\frac{1-27}{2},\frac{1+27}{2}\Rightarrow x=-13,14.\end{array}$

The difference of a number whose square with the same number which is equal to $182$ are $14\&-13$ .