Question

solve for x: abx^2+(b^2-ac)x-bc=0

Answer 1

x = -b/a, c/b

Step-by-step explanation:

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

or, $abx^2+b^2x-acx-bc=0$

or, $bx(ax+b)-c(ax+b)=0$

or, $(ax+b)(bx-c)=0$

$\impies x=-\frac{b}{a}, x=\frac{c}{b}$

Hope this answer is helpful.

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Answer 2

Answer:

Answer:

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

Step-by-step explanation:

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