Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 10 - Quadratic Equations - 10.3 - Quadratic Formula - Problem Set 10.3: 61

Answer

{$-1.95,2.15$}

Work Step by Step

Step 1: Comparing $-5x^{2}+x+21=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$, we obtain: $a=-5$, $b=1$ and $c=21$ Step 2: The quadratic formula is: $x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$ Step 3: Substituting the values of a, b and c in the formula: $x=\frac{-(1) \pm \sqrt {(1)^{2}-4(-5)(21)}}{2(-5)}$ Step 4: $x=\frac{-1 \pm \sqrt {1+420}}{-10}$ Step 5: $x=\frac{-1 \pm \sqrt {421}}{-10}$ Step 6: $x=\frac{-1 \pm 20.518}{-10}$ Step 7: $x=\frac{-1 + 20.518}{-10}$ or $x=\frac{-1 - 20.518}{-10}$ Step 8: $x=\frac{19.518}{-10}$ or $x=\frac{-21.518}{-10}$ Step 9: $x=-1.9518\approx-1.95$ or $x=2.1518\approx2.15$ Step 10: Therefore, the solution set is {$-1.95,2.15$}.
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