College Algebra (6th Edition)

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

Chapter 4 - Exponential and Logarithmic Functions - Exercise Set 4.3 - Page 478: 135


The solutions are $x=\dfrac{7}{2}\pm\dfrac{\sqrt{61}}{2}$

Work Step by Step

$x(x-7)=3$ Evaluate the product on the left side of the equation: $x^{2}-7x=3$ Take $3$ to the left side: $x^{2}-7x-3=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ In this case, $a=1$, $b=-7$ and $c=-3$ Substitute the known values into the formula and evaluate: $x=\dfrac{-(-7)\pm\sqrt{(-7)^{2}-4(1)(-3)}}{2(1)}=\dfrac{7\pm\sqrt{49+12}}{2}=...$ $...=\dfrac{7\pm\sqrt{61}}{2}=\dfrac{7}{2}\pm\dfrac{\sqrt{61}}{2}$
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