Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.4 Quadratic Equations - 1.4 Exercises - Page 122: 53


$\color{blue}{\left\{3-\sqrt2, 3+\sqrt2\right\}}$

Work Step by Step

Add $7$ to both sides of the equation to obtain: $x^2-6x+7=0$ RECALL: The quadratic equation $ax^2+bx+c=0$ can be solved using the quadratic formula: $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ The given equation has: $a=1, b=-6, c=7$ Substitute these values into the quadratic formula to obtain: $x=\dfrac{-(-6) \pm \sqrt{(-6)^2-4(1)(7)}}{2(1)} \\x=\dfrac{6\pm\sqrt{36-28}}{2} \\x=\dfrac{6\pm\sqrt{8}}{2} \\x=\dfrac{6\pm\sqrt{4(2)}}{2} \\x=\dfrac{6\pm2\sqrt{2}}{2} \\x=3 \pm \sqrt2$ Thus, the solution set is $\color{blue}{\left\{3-\sqrt2, 3+\sqrt2\right\}}$.
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