Elementary Algebra

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

Chapter 11 - Additional Topics - Chapter 11 Review Problem Set - Page 520: 65


{$\frac{-1 - i\sqrt {47}}{8},\frac{-1 + i\sqrt {47}}{8}$}

Work Step by Step

Step 1: First, we need to transform $x(4x+1)=-3$ into the standard form of a quadratic equation: $ax^{2}+bx+c=0$: $x(4x+1)=-3$ $4x^{2}+x+3=0$ Comparing $4x^{2}+x+3=0$ to the standard form of the quadratic equation, $ax^{2}+bx+c=0$, we find: $a=4$, $b=1$ 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: $x=\frac{-(1) \pm \sqrt {(1)^{2}-4(4)(3)}}{2(4)}$ Step 4: $x=\frac{-1 \pm \sqrt {1-48}}{8}$ Step 5: $x=\frac{-1 \pm \sqrt {-47}}{8}$ Step 6: $x=\frac{-1 \pm \sqrt {-1\times47}}{8}$ Step 7: $x=\frac{-1 \pm (\sqrt {-1}\times\sqrt {47})}{8}$ Step 8: $x=\frac{-1 \pm (i\times \sqrt {47})}{8}$ Step 9: $x=\frac{-1 \pm i\sqrt {47}}{8}$ Step 10: $x=\frac{-1 - i\sqrt {47}}{8}$ or $x=\frac{-1 + i\sqrt {47}}{8}$ Step 11: Therefore, the solution set is {$\frac{-1 - i\sqrt {47}}{8},\frac{-1 + i\sqrt {47}}{8}$}.
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