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: 14


x=-$\frac{15}{2}$, 8

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 2$x^{2}$-x-120=0, then a=2, b=-1, and c=-120, and to solve for x, then all we must do is plug the numbers into the quadratic formula then simplify. $x=\frac{-(-1)\pm\sqrt{(-1)^2-4(2)(-120)}}{2(2)}$ =$\frac{1\pm\sqrt{1-(-960)}}{4}$ (-1 squared=1, 4*2*-120=-960, and 2*2=4) =$\frac{1\pm\sqrt{961}}{4}$ (1-(-960)=961) =$\frac{1\pm31}{4}$ (The square root of 961=31) Since there is a plus minus symbol, we of course need to make sure that we compute both of our answers. x=$\frac{1+31}{4}$=$\frac{32}{4}$=8 x=$\frac{1-31}{4}$=$\frac{-30}{4}$=-$\frac{15}{2}$
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