Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.5 - Equations - 1.5 Exercises - Page 56: 57



Work Step by Step

$x^{2}+2x-5=0$ Since we need to solve this by completing the square, let's take the independent term to the right side of the equation. If we do that, the equation becomes: $x^{2}+2x=5$ We remember, that in order to complete this square, we add $(\dfrac{b}{2})^{2}$ to both sides of the equation. Also remember, that $b$ is always the coefficient of the first degree term. For this equation, $b=2$. $x^{2}+2x+(\dfrac{2}{2})^{2}=5+(\dfrac{2}{2})^{2}$ $x^{2}+2x+1=6$ On the left side of the equation, we are left with a perfect square trinomial. So we factor it and the equation becomes: $(x+1)^{2}=6$ Take the square root of both sides: $\sqrt{(x+1)^{2}}=\sqrt{6}$ $x+1=\pm\sqrt{6}$ Solve for $x$: $x=-1\pm\sqrt{6}$
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.