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: 49

Answer

The solutions are $x=-4$ and $x=\dfrac{1}{2}$

Work Step by Step

$2x^{2}+7x=4$ Take out common factor $2$ from the left side of the equation: $2\Big(x^{2}+\dfrac{7}{2}x\Big)=4$ Take $2$ to divide the right side: $x^{2}+\dfrac{7}{2}x=\dfrac{4}{2}$ $x^{2}+\dfrac{7}{2}x=2$ Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. In this particular case, $b=\dfrac{7}{2}$: $x^{2}+\dfrac{7}{2}x+\Big(\dfrac{7}{2\cdot2}\Big)^{2}=2+\Big(\dfrac{7}{2\cdot2}\Big)^{2}$ $x^{2}+\dfrac{7}{2}x+\dfrac{49}{16}=2+\dfrac{49}{16}$ $x^{2}+\dfrac{7}{2}x+\dfrac{49}{16}=\dfrac{81}{16}$ Factor the left side of the equation, which is a perfect square trinomial: $\Big(x+\dfrac{7}{4}\Big)^{2}=\dfrac{81}{16}$ Take the square root of both sides: $\sqrt{\Big(x+\dfrac{7}{4}\Big)^{2}}=\pm\sqrt{\dfrac{81}{16}}$ $x+\dfrac{7}{4}=\pm\dfrac{9}{4}$ Solve for $x$: $x=\pm\dfrac{9}{4}-\dfrac{7}{4}$ $x=\dfrac{9}{4}-\dfrac{7}{4}=\dfrac{2}{4}=\dfrac{1}{2}$ $x=-\dfrac{9}{4}-\dfrac{7}{4}=-\dfrac{16}{4}=-4$ The solutions are $x=-4$ and $x=\dfrac{1}{2}$
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