Question

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

Answer 1

$x=\frac{-bc}{ad}$ Given as solution to

$p\left(x\right)=a{d}^2(\frac{ax}{b}+\frac{2c}{d})x+b{c}^2=0$

If $x=\frac{-bc}{ad}$ is a solution to Equation 1, then it must satisfy it and $p\left(\frac{-bc}{ad}\right)=0$

$p\left(\frac{-bc}{ad}\right)=a{d}^2\left[\frac{a}{b}\right(\frac{-bc}{ad})+\frac{2c}{d}]\left(\frac{bc}{ad}\right)+b{c}^2$

$p\left(\frac{-bc}{ad}\right)=d(\frac{-c}{d}+\frac{2c}{d})(-bc)+b{c}^2$

$=d\left(\frac{c}{d}\right)(-bc)+b{c}^2$

$=-b{c}^2+b{c}^2$

$=0$

$\Rightarrow x=\frac{-bc}{ad}$ is a root of $a{d}^2(\frac{ax}{b}+\frac{2c}{d})x+b{c}^2=0$