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: 52



Work Step by Step

$2y^{2}+12y+3=0$ Take the $3$ to the right side of the equation: $2y^{2}+12y=-3$ Take out common factor $2$ from the left side of the equation: $2(y^{2}+6y)=-3$ 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 case, $b=6$ $2\Big[y^{2}+6y+\Big(\dfrac{6}{2}\Big)^{2}\Big]=-3+2\Big(\dfrac{6}{2}\Big)^{2}$ $2(y^{2}+6y+9)=-3+18$ $2(y^{2}+6y+9)=15$ Factor the expression inside the parentheses, which is a perfect square trinomial: $2(y+3)^{2}=15$ Take the $2$ to divide the right side of the equation: $(y+3)^{2}=\dfrac{15}{2}$ Take the square root of both sides of the equation: $\sqrt{(y+3)^{2}}=\sqrt{\dfrac{15}{2}}$ $y+3=\pm\sqrt{\dfrac{15}{2}}$ Solve for $y$: $y=-3\pm\sqrt{\dfrac{15}{2}}$
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