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: 53


The required numbers are $4-2\sqrt{2}$ and $4+2\sqrt{2}$.

Work Step by Step

Let the required numbers are $a$ and $b$. Now, according to the question, we have $a+b=8$ and $a\cdot b=8$. Substitute $b=8-a$ from first equation to the second equation, to get $a(8-a)=8$ $\Rightarrow 8a-a^2=8$ $\Rightarrow a^2-8a+8=0$ $\Rightarrow a=\dfrac{-(-8)\pm \sqrt{(-8)^2-4(1)(8)}}{2}$ using quadratic formula $\Rightarrow a=\dfrac{8\pm \sqrt{64-32}}{2}$ $\Rightarrow a=\dfrac{8\pm \sqrt{32}}{2}$ $\Rightarrow a=\dfrac{8\pm 4\sqrt{2}}{2}$ $\Rightarrow a=4\pm 2\sqrt{2}$ If $a=4+2\sqrt{2}$, then, $b=8-a=4-2\sqrt{2}$ and when $a=4-2\sqrt{2}$, then, $b=8-a=4+2\sqrt{2}$. Therefore, the required numbers are $4+2\sqrt{2}$ and $4-2\sqrt{2}$.
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