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: 53


the measurements of the legs are $(2+2\sqrt{2})$ $cm$ each and the hypotenuse is $(4+2\sqrt{2})$ $cm$

Work Step by Step

Let $x$ be the legs of an isosceles right triangle. Then $x+2$ 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+2)^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)(2)+(2)^2 \\ x^2+x^2=x^2+4x+4 \\ (x^2+x^2-x^2)-4x-4=0 \\ x^2-4x-4=0 .\end{array} In the equation above, $a= 1 ,$ $b= -4 ,$ and $c= -4 .$ 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{-(-4)\pm\sqrt{(-4)^2-4(1)(-4)}}{2(1)} \\ x=\dfrac{4\pm\sqrt{16+16}}{2} \\ x=\dfrac{4\pm\sqrt{32}}{2} \\ x=\dfrac{4\pm\sqrt{16\cdot2}}{2} \\ x=\dfrac{4\pm\sqrt{(4)^2\cdot2}}{2} \\ x=\dfrac{4\pm4\sqrt{2}}{2} \\ x=\dfrac{\cancel2(2)\pm\cancel2(2)\sqrt{2}}{\cancel2} \\ x=2\pm2\sqrt{2} .\end{array} Since $x$ is a measurement (and hence should be nonnegtive), then $ x=2+2\sqrt{2} .$ Hence, the measurements of the legs, $x,$ are $(2+2\sqrt{2})$ $cm$ each and the hypotenuse, $x+2,$ is $(4+2\sqrt{2})$ $cm.$
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