Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.2 Linear and Quadratic Functions - Exercises - Page 19: 61

Answer

The prove is as shown below.

Work Step by Step

$\underline{\text{Direct Method:}}$ Let $\alpha$ and $\beta$ are roots for the polynomial $x^2+bx+c$, or, are solutions to the equation $x^2+bx+c=0$. Using the quadratic formula, the roots are $x=\dfrac{-b\pm\sqrt{b^2-4c}}{2}$. So, without loss of generality let $\alpha=\dfrac{-b+\sqrt{b^2-4c}}{2}$ and $\beta=\dfrac{-b-\sqrt{b^2-4c}}{2}$. Now, sum of roots $=\alpha +\beta$ $=\dfrac{-b+\sqrt{b^2-4c}}{2}+\dfrac{-b-\sqrt{b^2-4c}}{2}$ $=\dfrac{-b+\sqrt{b^2-4c}-b-\sqrt{b^2-4c}}{2}$ $=\dfrac{-2b}{2}$ $=-b$. $\Rightarrow b=-\alpha-\beta$. Product of roots $=\alpha \beta$ $=\dfrac{-b+\sqrt{b^2-4c}}{2}\cdot \dfrac{-b-\sqrt{b^2-4c}}{2}$ $=\dfrac{(-b)^2-(\sqrt{b^2-4c})^2}{4}$ using identity $(x-y)(x+y)=x^2-y^2$. $=\dfrac{b^2-(b^2-4c)}{4}$ $=\dfrac{b^2-b^2+4c}{4}$ $=\dfrac{4c}{4}$ $=c$ Therefore, $b=-\alpha-\beta$ and $c=\alpha \beta$. hence, proved. $\underline{\text{Alternative Method:}}$ Let $\alpha$ and $\beta$ are roots for the polynomial $x^2+bx+c$, or, are solutions to the equation $x^2+bx+c=0$. Thus, we have $\alpha^2+b\alpha+c=0$ and $\beta^2+b\beta+c=0$. On subtracting both equations, we get $(\alpha^2+b\alpha+c)-(\beta^2+b\beta+c)=0$ $\alpha^2-\beta^2+b\alpha-b\beta=0$ $(\alpha-\beta)(\alpha+\beta)+b(\alpha-\beta)=0$ $(\alpha+\beta)+b=0$, or, $b=-\alpha-\beta$. Substitute this value in first equation, to get $\alpha^2+(-\alpha-\beta)\alpha+c=0$ $\Rightarrow \alpha^2-\alpha^2-\alpha\beta+c=0$ $-\alpha\beta+c=0$, or, $c=\alpha\beta$. Therefore, $b=-\alpha-\beta$ and $c=\alpha\beta$. Hence, proved.
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