Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.5 - Equations - 1.5 Exercises - Page 57: 86

Answer

This equation has two distinct real solutions.

Work Step by Step

$x^{2}+rx-s=0$ $(s\gt0)$ The formula for the discriminant is $D=b^{2}=4ac$. For this equation, $a=1$, $b=r$ and $c=-s$ Substitute the known values in the formula: $D=r^{2}-4(1)(-s)=r^{2}+4s$ We can be sure $r^{2}$ is a positive number, because any real number squared produces a positive number, and since we know $s\gt0$, then we can be certain that $r^{2}+4s$ is also a positive number. Knowing these facts, we can say that $D\gt0$ and that means that this equation has two distinct real solutions.
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