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: 59



Work Step by Step

$x^{2}-6x-11=0$ Let's take the independent term to the right side of the equation: $x^{2}-6x=11$ Let's remember that, in order to complete the square, we need to add $(\dfrac{b}{2})^{2}$ to both sides of the equation. $b$ is always the coefficient of the first degree term. In this equation, $b=-6$. $x^{2}-6x+(\dfrac{-6}{2})^{2}=11+(\dfrac{-6}{2})^{2}$ $x^{2}-6x+9=11+9$ $x^{2}-6x+9=20$ We have a perfect square trinomial on the left side of the equation. We factor it and the equation becomes: $(x-3)^{2}=20$ Take the square root of both terms: $\sqrt{(x-3)^{2}}=\sqrt{20}$ $x-3=\pm\sqrt{20}$ $x=3\pm\sqrt{20}$ Since $\sqrt{20}=\sqrt{(4)(5)}=2\sqrt{5}$, we can also write the solution like this: $x=3\pm2\sqrt{5}$
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