Question

${abx}^{2} + ( {b}^{2} - ac)x - bc = 0$
solve for x​

Answer 1

The equation we have

${abx}^{2} + ( {b}^{2} - ac)x - bc = 0$

After observing the equation we came to know that this equation is quadratic equation

so Using quadratic formula

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

Here,

a=ab

b=b^2-ac

c=bc

First we find d(discriminant)to find real roots or not then, we solve the question

$d = ( {b - ac})^{2} - 4(ab)( - bc) \\ d = {b}^{4} - 2a {b}^{2} c + {a}^{2} {c}^{2} + 4a {b}^{2} c \\ d = {b}^{4} + {a}^{2} {c}^{2} + 2a {b}^{2}c$

Now using identity made equation small

$d = {( {b}^{2} - ac) }^{2}$

Now we start finding the value of x

$x = \frac{- ( {b}^{2} - ac)± \sqrt{( { { {b}^{2} + ac)} }^{2} }}{2ab} \\ => x = \frac{ - {b}^{2} + ac+ {b}^{2} + ac }{2ab} or \frac{ - {b}^{2} + ac - {b}^{2} - ac }{2ab} \\ = > x = \frac{2ac}{2ab} or \: x = \frac{ - 2 {b}^{2} }{2ab} = > x = \frac{c}{b}or \frac{ - b}{a}$