Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 8 - Section 8.1 - Solving Quadratic Equations by Completing the Square - Exercise Set - Page 485: 74

Answer

$x=\left\{ \dfrac{15-7\sqrt{5}}{10},\dfrac{15+7\sqrt{5}}{10} \right\}$

Work Step by Step

Using the completing the square method, the solutions of the given quadratic equation, $ 10y^2-30y=2 ,$ is \begin{array}{l}\require{cancel} \dfrac{1}{10}\cdot (10y^2-30y)=(2)\cdot\dfrac{1}{10} \\\\ y^2-3y=\dfrac{1}{5} \\\\ y^2-3y+\left( \dfrac{-3}{2} \right)^2=\dfrac{1}{5}+\left( \dfrac{-3}{2} \right)^2 \\\\ y^2-3y+\dfrac{9}{4}=\dfrac{1}{5}+\dfrac{9}{4} \\\\ \left( y-\dfrac{3}{2} \right)^2=\dfrac{4}{20}+\dfrac{45}{20} \\\\ \left( y-\dfrac{3}{2} \right)^2=\dfrac{49}{20} \\\\ y-\dfrac{3}{2}=\pm\sqrt{\dfrac{49}{20}} \\\\ y-\dfrac{3}{2}=\pm\sqrt{\dfrac{49}{20}\cdot\dfrac{5}{5}} \\\\ y-\dfrac{3}{2}=\pm\sqrt{\dfrac{49}{100}\cdot5} \\\\ y-\dfrac{3}{2}=\pm\dfrac{7}{10}\sqrt{5} \\\\ y-\dfrac{3}{2}=\pm\dfrac{7\sqrt{5}}{10} \\\\ y=\dfrac{3}{2}\pm\dfrac{7\sqrt{5}}{10} \\\\ y=\dfrac{15}{10}\pm\dfrac{7\sqrt{5}}{10} \\\\ y=\dfrac{15\pm7\sqrt{5}}{10} .\end{array} Hence, $ x=\left\{ \dfrac{15-7\sqrt{5}}{10},\dfrac{15+7\sqrt{5}}{10} \right\} .$
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