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: 52



Work Step by Step

Step 1: Comparing $x^{2}-7x-13=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain: $a=1$, $b=-7$, and $c=-13$ 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: $x=\frac{-(-7) \pm \sqrt {(-7)^{2}-4(1)(-13)}}{2(1)}$ Step 4: $x=\frac{7 \pm \sqrt {49+52}}{2}$ Step 5: $x=\frac{7 \pm \sqrt {101}}{2}$ Step 6: $x=\frac{7 \pm 10.050}{2}$ Step 7: $x=\frac{7+10.050}{2}$ or $x=\frac{7-10.050}{2}$ Step 8: $x=\frac{17.050}{2}$ or $x=\frac{-3.050}{2}$ Step 9: $x\approx8.52$ or $x=\approx-1.52$ Step 10: Therefore, the solution set is {$-1.52,8.52$}.
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