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: 56


The solutions are $x=1\pm3i$

Work Step by Step

$x^{2}=2x-10$ Take all terms to the left side of the equation: $x^{2}-2x+10=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. In this case, $a=1$, $b=-2$ and $c=10$ Substitute the known values into the formula and evaluate: $x=\dfrac{-(-2)\pm\sqrt{(-2)^{2}-4(1)(10)}}{2(1)}=\dfrac{2\pm\sqrt{4-40}}{2}=...$ $...=\dfrac{2\pm\sqrt{-36}}{2}=\dfrac{2\pm6i}{2}=\dfrac{2}{2}\pm\dfrac{6}{2}i=1\pm3i$ The solutions are $x=1\pm3i$
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