Chapter 7.7, Problem 39WE

To determine

Toshow: that the root of $a{x}^2+bx+c=0$ are reciprocals of each other if and only if $a=c$ .

Explanation of Solution

Given information:

The quadratic equation is $a{x}^2+bx+c=0$ .

Proof: the roots of the equation can be find as:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Since, both roots are reciprocal to each other,

$\begin{array}{l}\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{2a}{-b-\sqrt{b^2-4ac}}\\ b^2-\left(b^2-4ac\right)=4a^2\text{ [cross multiply]}\\ 4ac=4a^2\text{ [subtract]}\\ a=c\text{ [simplify]}\end{array}$

Thus, the root of $a{x}^2+bx+c=0$ are reciprocals of each other if and only if $a=c$ .