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: 38

Answer

$x=5$

Work Step by Step

$\dfrac{2}{3}x^{2}-\dfrac{20}{3}x=-\dfrac{100}{6}$ Multiply the whole equation by $6$ to avoid working with fractions: $6\Big(\dfrac{2}{3}x^{2}-\dfrac{20}{3}x=-\dfrac{100}{6}\Big)$ $4x^{2}-40x=-100$ Take all terms to the left side: $4x^{2}-40x+100=0$ Take out common factor $4$ from the left side of the equation: $4(x^{2}-10x+25)=0$ Take the common factor to multiply the right side of the equation. Since that side is $0$, the result is the expression inside the parentheses equal to $0$ $x^{2}-10x+25=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a=1$, $b=-10$ and $c=25$ Substitute: $x=\dfrac{-(-10)\pm\sqrt{(-10)^{2}-4(1)(25)}}{2(1)}=\dfrac{10\pm\sqrt{100-100}}{2}=...$ $...=\dfrac{10\pm\sqrt{0}}{2}=5$
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