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: 65

Answer

$x=-\dfrac{1}{2}\pm\sqrt{\dfrac{23}{12}}$

Work Step by Step

$3x^{2}+3x=5$ Take out common factor $3$ from the left side of the equation: $3(x^{2}+x)=5$ Take the $3$ to divide the right side of the equation: $x^{2}+x=\dfrac{5}{3}$ Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. In this case, $b=1$ $x^{2}+x+\Big(\dfrac{1}{2}\Big)^{2}=\dfrac{5}{3}+\Big(\dfrac{1}{2}\Big)^{2}$ $x^{2}+x+\dfrac{1}{4}=\dfrac{5}{3}+\dfrac{1}{4}$ $x^{2}+x+\dfrac{1}{4}=\dfrac{23}{12}$ Factor the expression on the left side of the equation, which is a perfect square trinomial: $\Big(x+\dfrac{1}{2}\Big)^{2}=\dfrac{23}{12}$ Take the square root of both sides of the equation: $\sqrt{\Big(x+\dfrac{1}{2}\Big)^{2}}=\sqrt{\dfrac{23}{12}}$ $x+\dfrac{1}{2}=\pm\sqrt{\dfrac{23}{12}}$ Solve for $x$: $x=-\dfrac{1}{2}\pm\sqrt{\dfrac{23}{12}}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.