College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 1, Equations and Graphs - Section 1.4 - Solving Quadratic Equations - 1.4 Exercises - Page 123: 69

Answer

The width is $25 ft$ and the length is $35 ft$.

Work Step by Step

$\underline{\textbf{Givens: }}$ $Area = 875 ft^2$ $Width = W $ $Length (L)= W+ 10ft$ We know Area of a Rectangle $= L \times W$ i.e $x(x+10)=875$ $\Rightarrow x(x+10)=875$ $\Rightarrow x^2+10x=875$ $\Rightarrow x^2+10x-875=0$ the quadratic formula, where $a=1, b=10$ and $c=-875$ $\Rightarrow x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(10)\pm\sqrt{(10)^2-4(1)(-875)}}{2(1)}$ $\Rightarrow x=\dfrac{-10\pm\sqrt{100+3500}}{2}=\dfrac{-10\pm\sqrt{3600}}{2}$ $\Rightarrow x=\dfrac{-10\pm60}{2}$ $\Rightarrow x=\dfrac{-10+60}{2}=\dfrac{50}{2}=25$ or $x=\dfrac{-10-60}{2}=\dfrac{-70}{2}=-35$ $\color{red}{(Rejected)} $ i.e $x=25$ Therefore, the width is $25 ft$ and the length is $35 ft$.
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