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

Answer

$\text{ad}^2\Big(\frac{\text{ax}}{\text{b}}+\frac{\text{2c}}{\text{d}}\Big)\text{x}+\text{bc}^2=0$By multiplying $ad^2x$ by $\frac{\text{ax}}{\text{b}}$ we get
$\text{ad}^2\text{x}\Big(\frac{\text{ax}}{\text{b}}\Big)$
$=\frac{\text{a}^2\text{d}^2\text{x}^2}{ \text{b}}$
By multiplying $ad^2x$ by $\frac{\text{2c}}{\text{d}}$ we get
$\text{ad}^2\text{x}\Big(\frac{\text{2c}}{\text{d}}\Big)$
$=\frac{\text{2ca}\text{d}^2\text{x}}{\text{d}}$
By following the equation
$\frac{\text{a}^2\text{d}^2\text{x}^2}{\text{b}}+\frac{\text{2ca}\text{d}^2\text{x}}{\text{d}}+\text{bc}^2=0$
$\frac{\text{a}^2\text{d}^2\text{x}^2}{\text{b}}+\text{2cad}^{2-1}\text{x}+\text{bc}^2=0$
$\frac{\text{a}^2\text{d}^2\text{x}^2}{\text{b}}+\text{2cad}\text{x}+\text{bc}^2=0$
$a^2d^2x + 2abcdx + b^2c^2 = 0$ (by multiplying the denominator 'b' to 2cadx & $bc^2$​​​​​​​)
$\Rightarrow (adx + bc) = 0$ (Squaring)
$\Rightarrow adx + bc = 0$
$\Rightarrow adx = -bc$
$\Rightarrow\text{x}=\frac{-\text{bc}}{\text{ad}}$

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