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: 87

Answer

The solution set is $\left\{-\dfrac{-2-\sqrt2}{2}, \dfrac{2-\sqrt2}{2}\right\}$.

Work Step by Step

Subtract $-\frac{1}{2}$ on both sides of the equation to obtain: $x^2+\sqrt{2}x-\frac{1}{2}=0$ RECALL: A quadratic equation of the form $ax^2+bx+c=0$ can be solved using the quadratic formula $x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ The given quadratic equation has: $a=1 \\b=\sqrt{2} \\c=-\frac{1}{2}$ Substitute these values into the quadratic formula to obtain: $z=\dfrac{-\sqrt2 \pm \sqrt{(\sqrt2)^2-4(1)(-\frac{1}{2})}}{2(1)} \\z=\dfrac{-\sqrt2 \pm \sqrt{2-(-2)}}{2} \\z=\dfrac{-\sqrt2\pm \sqrt{2+2}}{2} \\x =\dfrac{-\sqrt2\pm\sqrt{4}}{2} \\x=\dfrac{-\sqrt2\pm\sqrt{2^2}}{2} \\x=\dfrac{-\sqrt2\pm 2}{2}$ Split the solutions to obtain: $x_1 = \dfrac{-\sqrt2+2}{2}=\dfrac{2-\sqrt{2}}{2} \\x_2=\dfrac{-\sqrt2-2}{2}=\dfrac{-2-\sqrt2}{2}$ Therefore, the solution set is $\left\{-\dfrac{-2-\sqrt2}{2}, \dfrac{2-\sqrt2}{2}\right\}$.
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