Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 8 - Section 8.2 - Solving Quadratic Equations by the Quadratic Formula - Exercise Set - Page 493: 54


the measurements of the legs are $(1+\sqrt{2})$ $m$ each and the hypotenuse is $(2+\sqrt{2})$ $m.$

Work Step by Step

Let $x$ be the legs of an isosceles right triangle. Then $x+1$ is the length of the hypotenuse. Using $a^2+b^2=c^2$ or the Pythagorean Theorem, then \begin{array}{l}\require{cancel} x^2+x^2=(x+1)^2 .\end{array} Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the equation above is equivalent to \begin{array}{l}\require{cancel} x^2+x^2=(x)^2+2(x)(1)+(1)^2 \\ x^2+x^2=x^2+2x+1 \\ (x^2+x^2-x^2)-2x-1=0 \\ x^2-2x-1=0 .\end{array} In the equation above, $a= 1 ,$ $b= -2 ,$ and $c= -1 .$ Using the Quadratic Formula which is given by $ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} ,$ then \begin{array}{l}\require{cancel} x=\dfrac{-(-2)\pm\sqrt{(-2)^2-4(1)(-1)}}{2(1)} \\ x=\dfrac{2\pm\sqrt{4+4}}{2} \\ x=\dfrac{2\pm\sqrt{8}}{2} \\ x=\dfrac{2\pm\sqrt{4\cdot2}}{2} \\ x=\dfrac{2\pm\sqrt{(2)^2\cdot2}}{2} \\ x=\dfrac{2\pm2\sqrt{2}}{2} \\ x=\dfrac{\cancel2\pm\cancel2\sqrt{2}}{\cancel2} \\ x=1\pm\sqrt{2} .\end{array} Since $x$ is a measurement (and hence should be nonnegative), then $ x=1+\sqrt{2} .$ Hence, the measurements of the legs, $x,$ are $(1+\sqrt{2})$ $m$ each and the hypotenuse, $x+1,$ is $(2+\sqrt{2})$ $m.$
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