Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 9 - Quadratic Functions and Equations - 9-6 The Quadratic Formula and the Discriminant - Practice and Problem-Solving Exercises - Page 571: 11


x=-$\frac{15}{18}$, $\frac{10}{3}$

Work Step by Step

The quadratic formula states that if a$x^{2}$+bx+c=0, and a$\ne$0, then $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. If 18$x^{2}$-45x-50=0, then a=18, b=-45, and c=-50, and to solve for x, then all we must do is plug the numbers into the quadratic formula then simplify. $x=\frac{-(-45)\pm\sqrt{(-45)^2-4(18)(-50)}}{2(18)}$ =$\frac{45\pm\sqrt{2025-(-3600)}}{36}$ (-45 squared=2025, 4*18*-50=-3600, and 2*18=36) =$\frac{45\pm\sqrt{5625}}{36}$ (2025-(-3600)=5625) =$\frac{45\pm75}{36}$ (The square root of 5625=75) Since there is a plus minus symbol, we of course need to make sure that we compute both of our answers. x=$\frac{45+75}{36}$=$\frac{120}{36}$=$\frac{10}{3}$ x=$\frac{45-75}{36}$=$\frac{-30}{36}$=-$\frac{15}{18}$
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