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Question

Using quadratic formula, solve the following equation for x :
$$abx^2+(b^2-ac)x-bc=0$$


Solution

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

$$\therefore x=\dfrac{-(b^2-ac)\pm \sqrt{(b^2-ac)^2-4(ab)(-bc)}}{2ab}$$

$$ x=\dfrac{-(b^2-ac)\pm \sqrt{(b^2-ac)^2+4ab^2c}}{2ab}$$

$$ x=\dfrac{-(b^2-ac)\pm \sqrt{(b^2+ac)^2}}{2ab}$$

$$x=\dfrac{-(b^2-ac)+(b^2+ac)}{2ab}, x=\dfrac{-(b^2-ac)-(b^2+ac)}{2ab}$$

$$x=\dfrac{2ac}{2ab}, x=\dfrac{-2b^2}{2ab}\Rightarrow x=\dfrac{c}{b}, x=\dfrac{-b}{a}$$ 

Mathematics

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