Question

factorize abx^2 +(b^2-ac)x - bc=0​

Answer 1

Step-by-step explanation:

abx²+(b²-ac) x-bc=0

b²-4ac

(b²-ac)²-4(ab)(-bc) =0

b*4+ac²-2b²ac+4ab²c=0

b*4+ac²+2ab²c

=(b²+ac)²

By formula method

x=-(b²-ac)±root(b²+AC)²/2ab

x=-b²+ac±b²+ac/2ab

x=-b²+ac+b²+ac/2ab,, ,x=-b²+ac-b²+ac/2ab

x=2ac/2ab=c/b. , x=-2b²/2ab=b/a

Answer 2

Step-by-step explanation:

Answer:

$\boxed{\mathfrak{x = - \frac{b}{a} \: \: \: \: or \: \: \: \: x = \frac{c}{b}}}$

Step-by-step explanation:

$\sf Solve \: for \: x: \\ \sf \implies ab {x}^{2} + ( {b}^{2} - ac)x - bc = 0 \\ \\ \sf The \: left \: hand \: side \: factors \: into \: a \\ \sf product \: with \: two \: terms: \\ \sf \implies ab {x}^{2} + {b}^{2}x - acx - bc = 0 \\ \\ \sf \implies bx(ax + b) - c(ax + b) = 0 \\ \\ \sf \implies (ax + b)(bx - c) = 0 \\ \\ \sf Split \: into \: two \: equations \: and \: calculate \: x : \\ \sf \implies ax + b = 0 \: \: \: \: or \: \: \: \: bx - c = 0 \\ \\ \sf \implies ax = - b \: \: \: \: or \: \: \: \: bx = c \\ \\ \sf \implies x = - \frac{b}{a} \: \: \: \: or \: \: \: \: x = \frac{c}{b}$