Algebra: A Combined Approach (4th Edition)

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

Chapter 11 - Section 11.1 - Solving Quadratic Equations by Completing the Square - Exercise Set - Page 764: 48



Work Step by Step

$2x^{2}+14x-1=0$ Take the $-1$ to the right side of the equation: $2x^{2}+14x=1$ Take out common factor $2$ from the left side of the equation: $2(x^{2}+7x)=1$ Add $\Big(\dfrac{b}{2}\Big)^{2}$ to the expression inside the parentheses and $2\Big(\dfrac{b}{2}\Big)^{2}$ to the right side of the equation. For this particular equation, $b=7$ $2\Big[x^{2}+7x+\Big(\dfrac{7}{2}\Big)^{2}\Big]=1+2\Big(\dfrac{7}{2}\Big)^{2}$ $2\Big(x^{2}+7x+\dfrac{49}{4}\Big)=1+\dfrac{49}{2}$ Factor the polynomial inside the parentheses, which is a perfect square trinomial: $2\Big(x+\dfrac{7}{2}\Big)^{2}=\dfrac{51}{2}$ Take the $2$ to divide the right side of the equation: $\Big(x+\dfrac{7}{2}\Big)^{2}=\dfrac{51}{4}$ Take the square root of both sides of the equation: $\sqrt{\Big(x+\dfrac{7}{2}\Big)^{2}}=\pm\sqrt{\dfrac{51}{4}}$ $x+\dfrac{7}{2}=\pm\dfrac{\sqrt{51}}{2}$ Solve for $x$: $x=-\dfrac{7}{2}\pm\dfrac{\sqrt{51}}{2}$
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