College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Section 1.4 - Quadratic Equations - 1.4 Exercises - Page 112: 55

Answer

$x=\left\{ 1- 2i,1+ 2i \right\}$

Work Step by Step

Using the properties of equality, the given equation, $ x^2=2x-5 ,$ is equivalent to \begin{array}{l}\require{cancel} x^2-2x+5=0 .\end{array} The equation above has $a=1$, $b=-2$, and $c=5$. Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, the solutions to the equation above are \begin{array}{l}\require{cancel} x=\dfrac{-(-2)\pm\sqrt{(-2)^2-4(1)(5)}}{2(1)} \\\\ x=\dfrac{2\pm\sqrt{4-20}}{2} \\\\ x=\dfrac{2\pm\sqrt{-16}}{2} \\\\ x=\dfrac{2\pm\sqrt{-1}\cdot\sqrt{16}}{2} \\\\ x=\dfrac{2\pm i\sqrt{(4)^2}}{2} \\\\ x=\dfrac{2\pm 4i}{2} \\\\ x=\dfrac{2(1\pm 2i)}{2} \\\\ x=\dfrac{\cancel{2}(1\pm 2i)}{\cancel{2}} \\\\ x=1\pm 2i .\end{array} Hence, the solutions are $ x=\left\{ 1- 2i,1+ 2i \right\} .$
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