Question

Show that $x=-\frac{bc}{ad}$ is a solution of the quadratic equation $a{d}^2\left(\frac{ax}{b}+\frac{2c}{d}\right)x+b{c}^2=0$.

Answer 1

LHS;
Consider the quadratic equation;
$a{d}^2\left(\frac{ax}{b}+\frac{2c}{d}\right)x+b{c}^2=0$
Put $x=-\frac{bc}{ad}$ in the given equation.
$$\begin{gathered} a{d}^2\left[\frac{a\left(-{\displaystyle \frac{bc}{ad}}\right)}{b}+\frac{2c}{d}\right]\left(-\frac{bc}{ad}\right)+b{c}^2 \\ =\frac{a{d}^2}{bd}\left[ad\left(-\frac{bc}{ad}\right)+2bc\right]\left(-\frac{bc}{ad}\right)+b{c}^2 \\ =\frac{a{d}^2}{bd}\left[-bc+2bc\right]\left(-\frac{bc}{ad}\right)+b{c}^2 \\ =\frac{a{d}^2}{bd}\left(bc\right)\left(-\frac{bc}{ad}\right)+b{c}^2 \\ =-\frac{a{b}^2{c}^2{d}^2}{ab{d}^2}+b{c}^2 \\ =-b{c}^2+b{c}^2 \\ =0=\mathrm{RHS} \end{gathered}$$
Hence, $x=-\frac{bc}{ad}$ is a solution to the given quadratic equation.