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Question

Solve the following quadratic equation by factorization :
$$abx^2 \, + \, (b^2 \, - \, ac) \, x \, - \, bc \, = \, 0$$


A
ba,bc
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B
ba,cb
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C
ba,bc
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D
ba,cb
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Solution

The correct option is B $$\dfrac{-b}{a}, \, \dfrac{c}{b}$$
$$abx^2+(b^2-ac)x-bc=0$$
$$\Rightarrow$$  $$abx^2+b^2x-acx-bc=0$$
$$\Rightarrow$$  $$bx(ax+b)-c(ax+b)=0$$
$$\Rightarrow$$  $$(ax+b)(bx-c)=0$$
$$\Rightarrow$$  $$ax+b=0$$ and $$bx-c=0$$
$$\therefore$$  $$x=\dfrac{-b}{a}$$ and $$x=\dfrac{c}{b}$$

Mathematics

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