## Intermediate Algebra (12th Edition)

$p=\left\{ \dfrac{-4-\sqrt{91}}{3},\dfrac{-4+\sqrt{91}}{3} \right\}$
$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $p=\dfrac{5(5-p)}{3(p+1)} ,$ 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} p=\dfrac{5(5)+5(-p)}{3(p)+3(1)} \\\\ p=\dfrac{25-5p}{3p+3} .\end{array} Since $\dfrac{a}{b}=\dfrac{c}{d}$ implies $ad=bc$ or sometimes referred to as cross-multiplication, the equation above is equivalent to \begin{array}{l}\require{cancel} p(3p+3)=1(25-5p) \\\\ 3p^2+3p=25-5p .\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} 3p^2+(3p+5p)-25=0 \\\\ 3p^2+8p-25=0 .\end{array} The quadratic equation above has $a= 3 , b= 8 , c= -25 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} p=\dfrac{-8\pm\sqrt{8^2-4(3)(-25)}}{2(3)} \\\\ p=\dfrac{-8\pm\sqrt{64+300}}{6} \\\\ p=\dfrac{-8\pm\sqrt{364}}{6} .\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} p=\dfrac{-8\pm\sqrt{4\cdot91}}{6} \\\\ p=\dfrac{-8\pm\sqrt{(2)^2\cdot91}}{6} \\\\ p=\dfrac{-8\pm2\sqrt{91}}{6} .\end{array} Cancelling the common factors from all the terms results to \begin{array}{l}\require{cancel} p=\dfrac{\cancel2(-4)\pm\cancel2(1)\sqrt{91}}{\cancel2(3)} \\\\ p=\dfrac{-4\pm\sqrt{91}}{3} .\end{array} The solutions are \begin{array}{l}\require{cancel} p=\dfrac{-4-\sqrt{91}}{3} \\\\\text{OR}\\\\ p=\dfrac{-4+\sqrt{91}}{3} .\end{array} Hence, $p=\left\{ \dfrac{-4-\sqrt{91}}{3},\dfrac{-4+\sqrt{91}}{3} \right\} .$