Q. 99 - Precalculus Enhanced with Graphing Utilities

Question

If r1 and r2 are two solutions of a quadratic equation ax2 + bx + c = 0, it can be shown that

$r_1+r_2=-\frac{b}{a} and r_1{r}_2=\frac{c}{a}$

Solve this system of equations for r1 and r2.

Step-by-Step Solution

Answer

$$\begin{gathered} r_1=\frac{-b+\sqrt{b^2-4ac}}{2a} \\ {r}_2={\frac{-b-\sqrt{b^2-4ac}}{2a}}_{} \end{gathered}$$

Step 1. Given information

 r1 and r2 are two solutions of  a quadratic equation ax2 + bx + c = 0 

and $r_1+r_2=-\frac{b}{a} and r_1{r}_2=\frac{c}{a}$

Step 2. r1 in terms of r2

First we divide the equation $r_1{r}_2=\frac{c}{a}$ by r2 to express r1 in terms of r2.

$$\begin{gathered} \frac{r_1{r}_2}{r_2}=\frac{c}{a{r}_2} \\ {r}_1=\frac{c}{a{r}_2} \end{gathered}$$

Step 3. Calculations

Substitute $r_1=\frac{c}{a{r}_2}$ in $r_1+r_2=-\frac{b}{a}$ we get

$$\begin{gathered} \frac{c}{a{r}_2}+r_2=-\frac{b}{a} \\ a{r}_2\left(\frac{c}{a{r}_2}+r_2\right)=-\frac{b}{a}·a{r}_2 \\ c+a{r}_2^2=-b{r}_2 \\ a{r}_2^2+b{r}_2+c=0 \\ {r}_2=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \end{gathered}$$

Step 4. r2 in terms of r1

Divide the equation $r_1{r}_2=\frac{c}{a}$  by r1 to express r2 in terms of r1.

$$\begin{gathered} \frac{r_1{r}_2}{r_1}=\frac{c}{a{r}_1} \\ {r}_2=\frac{c}{a{r}_1} \end{gathered}$$

Substitute $r_2=\frac{c}{a{r}_1}$ in  $r_1+r_2=-\frac{b}{a}$ we get

$$\begin{gathered} \frac{c}{a{r}_1}+r_1=-\frac{b}{a} \\ c+a{r}_1^2=-b{r}_1 \\ a{r}_1^2+b{r}_1+c=0 \\ {r}_1=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \end{gathered}$$

Step 5. r1 and r2

$$\begin{gathered} r_1=\frac{-b+\sqrt{b^2-4ac}}{2a} \\ {r}_2=\frac{-b-\sqrt{b^2-4ac}}{2a} \end{gathered}$$