Find the zeros of the quafratic …

We have,
f(x) = abx² + (b² - ac)x - bc
= abx² + b²x - acx - bc
= bx(ax + b ) - c(ax + b)
= (ax + b) (bx - c)
Now r(x)=0 if 

ax+b=0 or bx-c=0

{tex}\style{font-family:Arial}{\begin{array}{l}\style{font-size:12px}{\mathrm i}\style{font-size:12px}.\style{font-size:12px}{\mathrm e}\style{font-size:12px}.\style{font-size:12px}\;\style{font-size:12px}{\mathrm x}\style{font-size:12px}=\style{font-size:12px}-\frac{\style{font-size:12px}{\mathrm b}}{\style{font-size:12px}{\mathrm a}}\style{font-size:12px}\;\style{font-size:12px}{\mathrm{or}}\style{font-size:12px}\;\style{font-size:12px}{\mathrm x}\style{font-size:12px}=\frac{\style{font-size:12px}{\mathrm c}}{\style{font-size:12px}{\mathrm b}}\\\end{array}}{/tex}

Thus, the zeroes of f(x) are : 

{tex}\style{font-family:Arial}{\style{font-size:12px}{\mathrm\alpha}\style{font-size:12px}=\style{font-size:12px}-\frac{\style{font-size:12px}{\mathrm b}}{\style{font-size:12px}{\mathrm a}}\style{font-size:12px}\;\style{font-size:12px}{\mathrm{and}}\style{font-size:12px}\;\style{font-size:12px}{\mathrm\beta}\style{font-size:12px}=\frac{\style{font-size:12px}{\mathrm c}}{\style{font-size:12px}{\mathrm b}}}{/tex}

{tex}\text{α+β=-}\frac{\mathrm b}{\mathrm a}\;+\frac{\mathrm c}{\mathrm b}=\frac{-\mathrm b^2+ac}{ab}=-\frac{\mathrm( b^2-ac)}{ab}--(1){/tex}

{tex}\text{αβ=}\frac{\mathrm b}{\mathrm a}\;\times-\frac{\mathrm c}{\mathrm b}=-\frac{\mathrm c}{\mathrm a}----(2){/tex}

Now for f(x) = abx² + (b² - ac)x - bc

A = ab , B= b² - ac, C = -b

{tex}-\frac BA=-\frac{b^2-ac}{ab}--(3){/tex}

{tex}\frac CA=\frac{-bc}{ab}=-\frac ca----(4){/tex}

From (1) & (3) and (2)  & (4)  we conclude:

{tex}\begin{array}{l}\text{α+β=-}\frac{\mathrm B}{\mathrm A}\\\end{array}{/tex}

{tex}\begin{array}{l}\text{αβ=}\frac C{\mathrm A}\\\end{array}{/tex}