College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.5 - Page 161: 101

Answer

The solutions are $x=2\pm\sqrt{3}i$

Work Step by Step

$x^{2}=4x-7$ Take all terms to the left side of the equation: $x^{2}-4x+7=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ In this case, $a=1$, $b=-4$ and $c=7$ Substitute the known values into the formula and evaluate: $x=\dfrac{-(-4)\pm\sqrt{(-4)^{2}-4(1)(7)}}{2(1)}=\dfrac{4\pm\sqrt{16-28}}{2}=...$ $...=\dfrac{4\pm\sqrt{-12}}{2}=\dfrac{4\pm2\sqrt{3}i}{2}=2\pm\sqrt{3}i$
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