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 - Page 102: 88

Answer

$$x = \sqrt 2 ± 2$$

Work Step by Step

The equation at hand is a quadratic equation; to solve it, we should re-write it to equal to zero: $$\frac{1}{2} x^{2} = \sqrt 2 x + 1$$$$\frac{1}{2} x^{2} - \sqrt2 x - 1 = 0$$An optional, but useful, next step is to allow for the leading coefficient of the equation to be $1$, thus simplifying our work a bit. This can be achieved by multiplying the entire equation by the leading coefficient's reciprocal number: $$(\frac{2}{1})(\frac{1}{2} x^{2} - \sqrt2 x - 1 = 0)$$$$x^{2} - 2\sqrt 2 x - 2 = 0$$ Now, because we have a square root coefficient, the most efficient way of solving the equation would definitely be going straight to the quadratic formula $x = \frac{-b ± \sqrt {b^{2}-4ac}}{2a}$, so we substitute as so: $$x = \frac{-(-2\sqrt{2}) ± \sqrt {(-2\sqrt{2})^{2}-4(1)(-2)}}{2(1)}$$$$x = \frac{2\sqrt{2} ± \sqrt {8 + 8}}{2}$$$$\frac{2\sqrt{2} ± \sqrt {16}}{2}$$$$\frac{2\sqrt{2} ± 4}{2}$$ And, by factoring out the 2, we can simplify the equation to give: $$x = \frac{(2)(\sqrt{2} ± 2)}{2} = \sqrt2 ± 2$$
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