Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 8 - Section 8.2 - The Quadratic Formula - 8.2 Exercises: 19


$t=\left\{ \dfrac{-2-\sqrt{10}}{2},\dfrac{-2+\sqrt{10}}{2} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $ -2t(t+2)=-3 ,$ express first in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula. $\bf{\text{Solution Details:}}$ Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} -2t(t)-2t(2)=-3 \\\\ -2t^2-4t=-3 .\end{array} Using the properties of equality, in the form $ax^2+bx+c=0,$ the expression above is equivalent to \begin{array}{l}\require{cancel} -2t^2-4t+3=0 \\\\ -1(-2t^2-4t+3)=-1(0) \\\\ 2t^2+4t-3=0 .\end{array} The quadratic equation above has $a= 2 , b= 4 , c= -3 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} t=\dfrac{-4\pm\sqrt{4^2-4(2)(-3)}}{2(2)} \\\\ t=\dfrac{-4\pm\sqrt{16+24}}{4} \\\\ t=\dfrac{-4\pm\sqrt{40}}{4} .\end{array} Simplifying the radical by writing the radicand as an expression that contains a factor that is a perfect power of the index and then extracting the root of that factor result to \begin{array}{l}\require{cancel} t=\dfrac{-4\pm\sqrt{4\cdot10}}{4} \\\\ t=\dfrac{-4\pm\sqrt{(2)^2\cdot10}}{4} \\\\ t=\dfrac{-4\pm2\sqrt{10}}{4} .\end{array} Cancelling the common factors from all the terms results to \begin{array}{l}\require{cancel} t=\dfrac{\cancel2(-2)\pm\cancel2(1)\sqrt{10}}{\cancel2(2)} \\\\ t=\dfrac{-2\pm\sqrt{10}}{2} .\end{array} The solutions are \begin{array}{l}\require{cancel} t=\dfrac{-2-\sqrt{10}}{2} \\\\\text{OR}\\\\ t=\dfrac{-2+\sqrt{10}}{2} .\end{array} Hence, $ t=\left\{ \dfrac{-2-\sqrt{10}}{2},\dfrac{-2+\sqrt{10}}{2} \right\} .$
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