Chapter 8 - Section 8.2 - The Quadratic Formula - 8.2 Exercises: 28

$x=\left\{ \dfrac{-5-\sqrt{41}}{8},\dfrac{-5+\sqrt{41}}{8} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $(2x-1)^2=x+2 ,$ express first in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula. $\bf{\text{Solution Details:}}$ 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} (2x)^2-2(2x)(1)+(1)^2=x+2 \\\\ 4x^2-4x+1=x+2 .\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} 4x^2+(-4x-x)+(1-2)=0 \\\\ 4x^2-5x-1=0 .\end{array} The quadratic equation above has $a= 4 , b= -5 , c= -1 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} x=\dfrac{-5\pm\sqrt{(-5)^2-4(4)(-1)}}{2(4)} \\\\ x=\dfrac{-5\pm\sqrt{25+16}}{8} \\\\ x=\dfrac{-5\pm\sqrt{41}}{8} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{-5-\sqrt{41}}{8} \\\\\text{OR}\\\\ x=\dfrac{-5+\sqrt{41}}{8} .\end{array} Hence, $x=\left\{ \dfrac{-5-\sqrt{41}}{8},\dfrac{-5+\sqrt{41}}{8} \right\} .$

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