Elementary Algebra

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

Chapter 10 - Quadratic Equations - 10.5 - Solving Problems Using Quadratic Equations - Problem Set 10.5 - Page 463: 11


-8 or 8

Work Step by Step

Let the unknown number be $x$. This means that one-half of the number is $\frac{x}{2}$. As a result, the square of the unknown number is $x^{2}$ whereas the square of one-half of the number is $(\frac{x}{2})^{2}$. Since the sum of the square of a number and the square of one-half of the number is 80, we write the following equation and solve it: $x^{2}+(\frac{x}{2})^{2}=80$ $x^{2}+(\frac{x^{2}}{4})=80$ $\frac{4x^{2}+x^{2}}{4}=80$ $\frac{5x^{2}}{4}=80$ $5x^{2}=320$ $x^{2}=\frac{320}{5}$ $x^{2}=64$ $x=\pm\sqrt {64}$ $x=\pm 8$ Check when x=8: $8^{2}+(\frac{8}{2})^{2}=80$ $8^{2}+(4)^{2}=80$ $64+16=80$ $80=80$ Check when x=-8: $(-8)^{2}+(\frac{-8}{2})^{2}=80$ $(-8)^{2}+(-4)^{2}=80$ $64+16=80$ $80=80$ Therefore, the unknown number can be either -8 or 8.
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