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 - Page 453: 24



Work Step by Step

Step 1: Comparing $4x^{2}-x-3=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain: $a=4$, $b=-1$, and $c=-3$ 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, we obtain: $x=\frac{-(-1) \pm \sqrt {(-1)^{2}-4(4)(-3)}}{2(4)}$ Step 4: $x=\frac{1 \pm \sqrt {1+48}}{8}$ Step 5: $x=\frac{1 \pm \sqrt {49}}{8}$ Step 6: $x=\frac{1 \pm 7}{8}$ Step 7: $x=\frac{1 + 7}{8}$ or $x=\frac{1 - 7}{8}$ Step 8: $x=\frac{8}{8}$ or $x=\frac{-6}{8}$ Step 9: $x=1$ or $x=\frac{-3}{4}$ Step 10: Therefore, the solution set is {$-\frac{3}{4},1$}.
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