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: 44


$x=(\frac{7+\sqrt 53}{2}, \frac{7-\sqrt 53}{2})$

Work Step by Step

Step-1 : Add $1$ on both sides of the equation $x^2-7x=1$ Step -2 : Add the square of half of the co-efficient of $x$ to both sides. Co-efficient of $x =-7$ Half of $-7 = \frac{1}{2}×-7=\frac{-7}{2}$ Square of $\frac{-7}{2}$ is $\frac{-7}{2}×\frac{-7}{2}=\frac{49}{4}$ The equation becomes $x^2-7x+\frac{49}{4}=1+\frac{49}{4}$ Step-3 Factor the trinomial and simplify the right hand side. $(x-\frac{7}{2})^2=\frac{4+49}{4}$ $(x-\frac{7}{2})^2=\frac{53}{4}$ Step-4 Use the square root property and solve for $x$ $(x-\frac{7}{2})=±\sqrt\frac{53}{4} $ Step-5 Add $\frac{7}{2}$ on both the sides $x=\frac{7}{2}±\sqrt \frac{53}{4}$ Step-6 Simplify the right hand side $x=\frac{7}{2}±\frac{\sqrt 53}{2}$ $x=\frac{7±\sqrt 53}{2}$ Therefore the solution set is $(\frac{7+\sqrt 53}{2}, \frac{7-\sqrt 53}{2})$
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