College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.2 - Quadratic Equations - 1.2 Assess Your Understanding: 38

Answer

The solution set is $\left\{3-\sqrt{22}, 3+\sqrt{22}\right\}$.

Work Step by Step

RECALL: A quadratic equation in the form $x^2 + bx =c$ can be solved by completing the square by adding $\left(\frac{b}{2}\right)^2$ to both sides of the equation then using the square root property. The equation will become $(x+\frac{b}{2})^2=c+(\frac{b}{2})^2$ Complete the square by adding $(\frac{-6}{2})^2=9$ to both sides of the equation to obtain: $x^2-6x+9=13+9 \\x^2-6x+9=22$ Write the trinomial as a square of a binomial to obtain: $(x-3)^2=22$ Take the square root of both sides to obtain: $x-3 = \pm\sqrt{22}$ Add $3$ to both sides to obtain: $x-3+3 = 3 \pm \sqrt{22} \\x = 3\pm \sqrt{22}$ Split the solutions to obtain: $x_1=3+\sqrt{22}$ $\\x_2=3-\sqrt{22}$ Thus, the solution set is $\left\{3-\sqrt{22}, 3+\sqrt{22}\right\}$.
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