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: 87

Answer

$x=100$ and $x=-50$

Work Step by Step

$\dfrac{x^{2}}{x+100}=50$ Take $x+100$ to multiply the right side of the equation: $x^{2}=50(x+100)$ Evaluate the product on the right: $x^{2}=50x+5000$ Rewrite the equation by taking all terms to the left side: $x^{2}-50x-5000=0$ Solve using the quadratic formula, knowing that $a=1$, $b=-50$ and $c=-5000$: $x=\dfrac{-(-50)\pm\sqrt{(-50)^{2}-4(1)(-5000)}}{2(1)}=...$ $...=\dfrac{50\pm\sqrt{2500+20000}}{2}=\dfrac{50\pm\sqrt{22500}}{2}=\dfrac{50\pm150}{2}$ So, our two solutions are: $x=\dfrac{50+150}{2}=100$ and $x=\dfrac{50-150}{2}=-50$
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