Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 11 - Section 11.2 - Solving Quadratic Equations by Completing the Square - Exercise Set - Page 775: 4


$x=3$ and $x=-\dfrac{1}{5}$

Work Step by Step

$5x^{2}-3=14x$ Take the $14x$ to the left side of the equation: $5x^{2}-14x-3=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For this equation, $a=5$, $b=-14$ and $c=-3$ Substitute: $x=\dfrac{-(-14)\pm\sqrt{(-14)^{2}-4(5)(-3)}}{2(5)}=\dfrac{14\pm\sqrt{196+60}}{10}=...$ $...=\dfrac{14\pm\sqrt{256}}{10}=\dfrac{14\pm16}{10}$ Our two solutions are: $x=\dfrac{14+16}{10}=3$ $x=\dfrac{14-16}{10}=-\dfrac{1}{5}$
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