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: 17

Answer

$x=\dfrac{3}{2}\pm\dfrac{\sqrt{11}}{2}$

Work Step by Step

$\dfrac{1}{3}y^{2}-y-\dfrac{1}{6}=0$ Multiply the whole equation by $6$ to avoid working with fractions: $6\Big(\dfrac{1}{3}y^{2}-y-\dfrac{1}{6}=0\Big)$ $2y^{2}-6y-1=0$ Use the quadratic formula to solve this equation. The formula is $y=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For this equation, $a=2$, $b=-6$ and $c=-1$ Substitute: $y=\dfrac{-(-6)\pm\sqrt{(-6)^{2}-4(2)(-1)}}{2(2)}=\dfrac{6\pm\sqrt{36+8}}{4}=...$ $...=\dfrac{6\pm\sqrt{44}}{4}=\dfrac{6\pm2\sqrt{11}}{4}=\dfrac{3}{2}\pm\dfrac{\sqrt{11}}{2}$
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