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: 62



Work Step by Step

$3x^{2}-6x-1=0$ Take the independent term to the right side of the equation: $3x^{2}-6x=1$ Take out common factor $3$ on the left side of the equation: $3(x^{2}-2x)=1$ Let's complete the square of the expression inside the parentheses. 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. For the expression inside the parentheses, $b=-2$. Please note that, since the expression whose square needs to be completed is multiplied by $3$, we will add $3(\dfrac{b}{2})^{2}$ to right side of the equation. $3[x^{2}-2x+(\dfrac{-2}{2})^{2}]=1+3(\dfrac{-2}{2})^{2}$ $3(x^{2}-2x+1)=1+3$ $3(x^{2}-2x+1)=4$ $x^{2}-2x+1=\dfrac{4}{3}$ We have a perfect square trinomial on the left side of the equation. We factor it and the equation becomes: $(x-1)^{2}=\dfrac{4}{3}$ Take the square root of both sides of the equation: $\sqrt{(x-1)^{2}}=\sqrt{\dfrac{4}{3}}$ $x-1=\pm\dfrac{2}{\sqrt{3}}$ Rationalizing the right side of the equation, and solving for $x$ we get: $x=1\pm\dfrac{2\sqrt{3}}{3}$
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