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: 65


The solution is $x=5$

Work Step by Step

$(x-9)(x-1)=-16$ Evaluate the product on the left side: $x^{2}-x-9x+9=-16$ Take $16$ to the left side and simplify: $x^{2}-x-9x+9+16=0$ $x^{2}-10x+25=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. In this case, $a=1$, $b=-10$ and $c=25$ Substitute the known values into the formula and evaluate: $x=\dfrac{-(-10)\pm\sqrt{(-10)^{2}-4(1)(25)}}{2(1)}=\dfrac{10\pm\sqrt{100-100}}{2}=...$ $...=\dfrac{10\pm\sqrt{0}}{2}=\dfrac{10}{2}=5$ The solution is $x=5$
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