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: 51



Work Step by Step

$3y^{2}+6y-4=0$ Take the $4$ to the right side of the equation: $3y^{2}+6y=4$ Take out common factor $3$ from the left side of the equation: $3(y^{2}+2y)=4$ Add $\Big(\dfrac{b}{2}\Big)^{2}$ to the expression inside the parentheses and $3\Big(\dfrac{b}{2}\Big)^{2}$ to the right side of the equation. For this particular case, $b=2$ $3\Big[y^{2}+2y+\Big(\dfrac{2}{2}\Big)^{2}\Big]=4+3\Big(\dfrac{2}{2}\Big)^{2}$ $3(y^{2}+2y+1)=4+3$ $3(y^{2}+2y+1)=7$ Factor the expression inside the parentheses, which is a perfect square trinomial: $3(y+1)^{2}=7$ Take the $3$ to divide the right side of the equation: $(y+1)^{2}=\dfrac{7}{3}$ Take the square root of both sides of the equation: $\sqrt{(y+1)^{2}}=\pm\sqrt{\dfrac{7}{3}}$ $y+1=\pm\sqrt{\dfrac{7}{3}}$ Solve for $y$: $y=-1\pm\sqrt{\dfrac{7}{3}}$
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