Question

HEYA HERE IS YOUR QUESTION,

solve this by using quadratic formula.

Answer 1

Given Equation is abx^2 + (b^2 - ac)x - bc = 0

On comparing with ax^2 + bx + c = 0,  we get

a = ab, b = b^2 - ac, c = -bc.

The solutions are:

$= \ \textgreater \ x = \frac{-b + \sqrt{b^2 - 4ac} }{2a}$

$= \ \textgreater \ \frac{-(b^2 - ac) + \sqrt{(b^2 - ac)^2 - 4(ab)(-bc)} }{2(ab)}$

$= \ \textgreater \ \frac{-(b^2 - ac)+ \sqrt{b^4 + a^2c^2 - 2b^2ac + 4b^2ac } }{2ab}$

$= \ \textgreater \ \frac{-(b^2 - ac) + \sqrt{b^4 + a^2c^2 + 2b^2ac} }{2ab}$

$= \ \textgreater \ \frac{-(b^2 - ac) + \sqrt{(b^2 + ac)^2} }{2ab}$

$= \ \textgreater \ \frac{-(b^2 - ac) + b^2 + ac}{2ab}$

$= \ \textgreater \ \frac{-b^2 + ac + b^2 + ac }{2ab}$

$= \ \textgreater \ \frac{2ac}{2ab}$

$= \ \textgreater \ \frac{c}{b}$


(2) 

$= \ \textgreater \ x = \frac{-b - \sqrt{b^2 - 4ac} }{2a}$

$= \ \textgreater \ \frac{-(b^2 - ac) - \sqrt{(b^2 - ac)^2 + 4ab * bc} }{2ab}$

$= \ \textgreater \ \frac{- (b^2 - ac) - \sqrt{b^4 + a^2c^2 - 2b^2ac + 4b^2ac} }{2ab}$

$= \ \textgreater \ \frac{-(b^2 - ac) - \sqrt{b^4 + a^2c^2 + 2b^2ac} }{2ab}$

$= \ \textgreater \ \frac{-(b^2 - ac) - \sqrt{(b^2 + ac)^2} }{2ab}$

$= \ \textgreater \ \frac{-(b^2 - ac) - b^2 + ac}{2ab}$

$= \ \textgreater \ \frac{-b^2 + ac - b^2 + ac }{2ab}$

$= \ \textgreater \ \frac{-2b^2}{2ab}$

$= \ \textgreater \ \frac{-b}{a}$



Hope this helps!

Answer 2

Hi,

Please see the attached file !



Thanks