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: 8


x=-6, $\frac{14}{5}$

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 5$x^{2}$+16x-84=0, then a=5, b=16, and c=-84, and to solve for x, then all we must do is plug the numbers into the quadratic formula then simplify. $x=\frac{-16\pm\sqrt{(16)^2-4(5)(-84)}}{2(5)}$ =$\frac{-16\pm\sqrt{256-(-1680)}}{10}$ (16 squared=256, 4*5*-84=-1680, and 2*5=10) =$\frac{-16\pm\sqrt{1936}}{10}$ (256-(-1680)=1936) =$\frac{-16\pm44}{10}$ (The square root of 1936=44) Since there is a plus minus symbol, we of course need to make sure that we compute both of our answers. x=$\frac{-16+44}{10}$=$\frac{28}{10}$=$\frac{14}{5}$ x=$\frac{-16-44}{10}$=$\frac{-60}{10}$=-6
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