Answer to Question #225915 in Chemical Engineering for Lokika

Question

The value of 1/D²+3D (2 + 6x) =

A) x

B) x² C) 0 D) 1

Answer 1

$D^2+3D\left(2+6x\right)=0\\ \mathrm{Subtract\:}D^2\mathrm{\:from\:both\:sides}\\ D^2+3D\left(2+6x\right)-D^2=0-D^2\\ \mathrm{Simplify}\\ 3D\left(2+6x\right)=-D^2\\ \mathrm{Divide\:both\:sides\:by\:}3D;\quad \:D\ne \:0\\ \frac{3D\left(2+6x\right)}{3D}=\frac{-D^2}{3D};\quad \:D\ne \:0\\ 2+6x=-\frac{D}{3};\quad \:D\ne \:0\\ \mathrm{Subtract\:}2\mathrm{\:from\:both\:sides}\\ 2+6x-2=-\frac{D}{3}-2;\quad \:D\ne \:0\\ \mathrm{Simplify}\\ 6x=-\frac{D}{3}-2;\quad \:D\ne \:0\\ \mathrm{Divide\:both\:sides\:by\:}6;\quad \:D\ne \:0\\ \frac{6x}{6}=-\frac{\frac{D}{3}}{6}-\frac{2}{6};\quad \:D\ne \:0\\ x=\frac{-D-6}{18};\quad \:D\ne \:0$