## Intermediate Algebra (12th Edition)

$x=\left\{ \dfrac{-3-\sqrt{57}}{8},\dfrac{-3+\sqrt{57}}{8} \right\}$
$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $(2x+1)^2=x+4 ,$ 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+4 \\\\ 4x^2+4x+1=x+4 .\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)+(1-4)=0 \\\\ 4x^2+3x-3=0 .\end{array} The quadratic equation above has $a= 4 , b= 3 , c= -3 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} x=\dfrac{-3\pm\sqrt{3^2-4(4)(-3)}}{2(4)} \\\\ x=\dfrac{-3\pm\sqrt{9+48}}{8} \\\\ x=\dfrac{-3\pm\sqrt{57}}{8} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{-3-\sqrt{57}}{8} \\\\\text{OR}\\\\ x=\dfrac{-3+\sqrt{57}}{8} .\end{array} Hence, $x=\left\{ \dfrac{-3-\sqrt{57}}{8},\dfrac{-3+\sqrt{57}}{8} \right\} .$