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: 20



Work Step by Step

Step 1: Comparing $x^{2}-15x+54=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain: $a=1$, $b=-15$, and $c=54$ 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: $x=\frac{-(-15) \pm \sqrt {(-15)^{2}-4(1)(54)}}{2(1)}$ Step 4: $x=\frac{15 \pm \sqrt {225-216}}{2}$ Step 5: $x=\frac{15 \pm \sqrt {9}}{2}$ Step 6: $x=\frac{15 \pm 3}{2}$ Step 7: $x=\frac{15+3}{2}$ or $x=\frac{15-3}{2}$ Step 8: $x=\frac{18}{2}$ or $x=\frac{12}{2}$ Step 9: $x=9$ or $x=6$ Step 10: Therefore, the solution set is {$6,9$}.
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