Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 8 - Section 8.3 - Quadratic Functions and Their Graphs - Exercise Set - Page 627: 62

Answer

The pair is $-12$ and $12$. The minimum product is $-144$.

Work Step by Step

The given difference of the numbers in the pair is $24$. Let's note by $x$ one of the numbers of the pair whose product is minimum. The other number is $x+24$. The product is $\Rightarrow P(x)=x(x+24)$ Use the distributive property. $\Rightarrow P(x)=x^2+24x$. The standard form of a quadratic equation is $f(x)=ax^2+bx+c$. Compare both equations to identify the constants $a=1$, $b=24$, $c=0$. The value of $a$ is greater than zero, so the function has the minimum value. The value of $x$ for which the minimum is reached is $-\frac{b}{2a}$. Substitute all values. $\Rightarrow x=-\frac{24}{2(1)}$. Simplify. $\Rightarrow x=-12$. The other number is $x+24=-12+24=12$. Hence, the pair is $-12$ and $12$. The minimum product is $12(-12)=-144$.
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