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: 50


Francesca was wrong because the quadratic formula works even when $b=0$.

Work Step by Step

RECALL: The quadratic equation $ax^2+bx+c=0$ can be solved using the quadratic formula: $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ The given equation has: $a=1, b=0, c=-19$ Substitute these values into the quadratic formula to obtain: $x=\dfrac{-0 \pm \sqrt{0^2-4(1)(-19)}}{2(1)} \\x=\dfrac{\pm\sqrt{0+76}}{2} \\x=\dfrac{\pm\sqrt{76}}{2} \\x=\dfrac{\pm\sqrt{4(19)}}{2} \\x=\dfrac{2\sqrt{19}}{2} \\x=\pm\sqrt{19}$ Thus, the solution set is $\color{blue}{\left\{-\sqrt{19}, \sqrt{19}\right\}}$. Francesca was wrong because the quadratic formula actually works regardless of whether $b=0$ or not.
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