Intermediate Algebra (12th Edition)

$x=\left\{ \dfrac{-2-i\sqrt{2}}{3},\dfrac{-2+i\sqrt{2}}{3} \right\}$
$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $x(3x+4)=-2 ,$ 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} x(3x)+x(4)=-2 \\\\ 3x^2+4x=-2 .\end{array} Using the properties of equality, the given equation is equivalent to \begin{array}{l}\require{cancel} 3x^2+4x+2=0 .\end{array} The quadratic equation above has $a= 3 , b= 4 , c= 2 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} x=\dfrac{-4\pm\sqrt{4^2-4(3)(2)}}{2(3)} \\\\ x=\dfrac{-4\pm\sqrt{16-24}}{6} \\\\ x=\dfrac{-4\pm\sqrt{-8}}{6} .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the equation above is equivalent to \begin{array}{l}\require{cancel} x=\dfrac{-4\pm\sqrt{-1}\cdot\sqrt{8}}{6} .\end{array} Since $i=\sqrt{-1},$ the equation above is equivalent to \begin{array}{l}\require{cancel} x=\dfrac{-4\pm i\sqrt{8}}{6} .\end{array} Simplifying the radicand by writing it as an expression that contains a factor that is a perfect square of the index and then extracting the root of that factor, the equation above is equivalent to \begin{array}{l}\require{cancel} x=\dfrac{-4\pm i\sqrt{4\cdot2}}{6} \\\\ x=\dfrac{-4\pm i\sqrt{(2)^2\cdot2}}{6} \\\\ x=\dfrac{-4\pm i(2)\sqrt{2}}{6} \\\\ x=\dfrac{-4\pm 2i\sqrt{2}}{6} .\end{array} Cancelling the common factor in each term results to \begin{array}{l}\require{cancel} x=\dfrac{\cancel2(-2)\pm \cancel2(1)i\sqrt{2}}{\cancel2(3)} \\\\ x=\dfrac{-2\pm i\sqrt{2}}{3} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{-2-i\sqrt{2}}{3} \\\\\text{OR}\\\\ x=\dfrac{-2+i\sqrt{2}}{3} .\end{array} Hence, $x=\left\{ \dfrac{-2-i\sqrt{2}}{3},\dfrac{-2+i\sqrt{2}}{3} \right\} .$