Chapter 3.4, Problem 73E

To determine

To prove: That the mean of the solutions is $-\frac{b}{2a}$ . Also, determine the relation of the fact to the symmetry of the graph of $y=a{x}^2+bx+c$ .

Answer to Problem 73E

The mean of the solutions is $-\frac{b}{2a}$ and the mean of the solutions and the symmetry are equal.

Explanation of Solution

Given information:

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

Calculation:

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

Consider the solutions of the given equation are $x_1$ and $x_2$ that is,

  $x_1=\frac{-b-\sqrt{b^2-4ac}}{2a}$ and $x_2=\frac{-b+\sqrt{b^2-4ac}}{2a}$ .

The mean of the above solutions is,

  $\begin{array}{c}\frac{x_1+x_2}{2}=\frac{\frac{-b-\sqrt{b^2-4ac}}{2a}+\frac{-b+\sqrt{b^2-4ac}}{2a}}{2}\\ \frac{x_1+x_2}{2}=\frac{\frac{-b-\sqrt{b^2-4ac}-b+\sqrt{b^2-4ac}}{2a}}{2}\\ \frac{x_1+x_2}{2}=\frac{\frac{-b-\cancel{\sqrt{b^2-4ac}}-b+\cancel{\sqrt{b^2-4ac}}}{2a}}{2}\\ \frac{x_1+x_2}{2}=\frac{\frac{-2b}{2a}}{2}\end{array}$

Simplify the above expression further.

  $\begin{array}{c}\frac{x_1+x_2}{2}=\frac{\frac{-\cancel{2}b}{\cancel{2}a}}{2}\\ =-\frac{b}{2a}\end{array}$

Rewrite the given equation in the vertex form as:

  $\begin{array}{c}y=a\left(x^2+\frac{b}{a}x\right)+c\\ y+a{\left(\frac{\frac{b}{a}}{2}\right)}^2=a\left(x^2+\frac{b}{a}x+{\left(\frac{\frac{b}{a}}{2}\right)}^2\right)\\ y=a{\left(x+\frac{b}{2a}\right)}^2+\frac{4ac}{4a}-\frac{b^2}{4a}\\ y=a{\left(x+\frac{b}{2a}\right)}^2+\frac{4ac-b^2}{4a}\end{array}$

The symmetry of the graph is $x=h$ ; this means the symmetry of the given equation is $x=-\frac{b}{2a}$

Hence, the mean of the solutions is $-\frac{b}{2a}$ and the mean of the solutions and the symmetry are equal.