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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 11: Q. 99 (page 775)

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

$r_{1} + r_{2} = - \frac{b}{a} a n d r_{1} r_{2} = \frac{c}{a}$
Solve this system of equations for r1 and r2.

Short Answer

Expert verified

$r_{1} = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a} r_{2} = {\frac{- b - \sqrt{b^{2} - 4 a c}}{2 a}}_{}$

Step by step solution

01

Step 1. Given information

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

and $r_{1} + r_{2} = - \frac{b}{a} a n d r_{1} r_{2} = \frac{c}{a}$

02

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.

$\frac{r_{1} r_{2}}{r_{2}} = \frac{c}{a r_{2}} r_{1} = \frac{c}{a r_{2}}$

03

Step 3. Calculations

Substitute $r_{1} = \frac{c}{a r_{2}}$ in $r_{1} + r_{2} = - \frac{b}{a}$ we get

role="math" localid="1646997829971" $\frac{c}{a r_{2}} + r_{2} = - \frac{b}{a} a r_{2} (\frac{c}{a r_{2}} + r_{2}) = - \frac{b}{a} 路 a r_{2} c + a r_{2}^{2} = - b r_{2} a r_{2}^{2} + b r_{2} + c = 0 r_{2} = \frac{- b 卤 \sqrt{b^{2} - 4 a c}}{2 a}$

04

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.

$\frac{r_{1} r_{2}}{r_{1}} = \frac{c}{a r_{1}} r_{2} = \frac{c}{a r_{1}}$

Substitute $r_{2} = \frac{c}{a r_{1}}$ in $r_{1} + r_{2} = - \frac{b}{a}$ we get

$\frac{c}{a r_{1}} + r_{1} = - \frac{b}{a} c + a r_{1}^{2} = - b r_{1} a r_{1}^{2} + b r_{1} + c = 0 r_{1} = \frac{- b 卤 \sqrt{b^{2} - 4 a c}}{2 a}$

05

Step 5. r1 and r2

$r_{1} = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a} r_{2} = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a}$

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