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Mathematics

Using quadratic formula find the value of x

abx2 + (b2 - ac)x - bc = 0

Quadratic Equations

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Answer

abx2 + (b2 - ac)x - bc = 0

Comparing abx2 + (b2 - ac)x - bc = 0 with ax2 + bx + c = 0 we get,

a = ab, b = (b2 - ac), c = -bc

We know that,

x = −b±b2−4ac2a\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting values we get,

⇒x=−(b2−ac)±(b2−ac)2−4(ab)(−bc)2ab=−b2+ac±a2c2+b4−2acb2+4acb22ab=−b2+ac±a2c2+b4+2acb22ab=−b2+ac±(ac+b2)22ab=−b2+ac±(ac+b2)2ab=−b2+ac+ac+b22ab or −b2+ac−(ac+b2)2ab=2ac2ab or −2b22ab=cb or −ba.\Rightarrow x = \dfrac{-(b^2 - ac) \pm \sqrt{(b^2 - ac)^2 - 4(ab)(-bc)}}{2ab} \\[1em] = \dfrac{-b^2 + ac \pm \sqrt{a^2c^2 + b^4 - 2acb^2 + 4acb^2}}{2ab} \\[1em] = \dfrac{-b^2 + ac \pm \sqrt{a^2c^2 + b^4 + 2acb^2}}{2ab} \\[1em] = \dfrac{-b^2 + ac \pm \sqrt{(ac + b^2)^2}}{2ab} \\[1em] = \dfrac{-b^2 + ac \pm (ac + b^2)}{2ab} \\[1em] = \dfrac{-b^2 + ac + ac + b^2}{2ab} \text{ or } \dfrac{-b^2 + ac - (ac + b^2)}{2ab} \\[1em] = \dfrac{2ac}{2ab} \text{ or } \dfrac{-2b^2}{2ab} \\[1em] = \dfrac{c}{b} \text{ or } -\dfrac{b}{a}.

Hence, x = cb,−ba.\dfrac{c}{b}, -\dfrac{b}{a}.

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