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: 47



Work Step by Step

$3p^{2}-12p+2=0$ Take the $2$ to substract to the right side of the equation: $3p^{2}-12p=-2$ Take out common factor $3$ from the left side of the equation: $3(p^{2}-4p)=-2$ 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 case, $b=-4$ $3\Big[p^{2}-4p+\Big(\dfrac{-4}{2}\Big)^{2}\Big]=-2+3\Big(\dfrac{-4}{2}\Big)^{2}$ $3(p^{2}-4p+4)=-2+12$ $3(p^{2}-4p+4)=10$ Now, factor the expression inside the parentheses, which is a perfect square trinomial: $3(p-2)^{2}=10$ Take the $3$ to divide to the right side of the equation: $(p-2)^{2}=\dfrac{10}{3}$ Take the square root of both sides of the equation: $\sqrt{(p-2)^{2}}=\pm\sqrt{\dfrac{10}{3}}$ $p-2=\pm\sqrt{\dfrac{10}{3}}$ Solve for $p$: $p=2\pm\sqrt{\dfrac{10}{3}}$
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