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Mathematics

Using quadratic formula find the value of x

p2x2 - (p2 - q2)x - q2 = 0

Quadratic Equations

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Answer

p2x2 - (p2 - q2)x - q2 = 0

Comparing p2x2 - (p2 - q2)x - q2 with ax2 + bx + c = 0 we get,

a = p2, b = -(p2 - q2) = (q2 - p2), c = -q2

We know that,

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

Substituting values we get,

⇒x=−b±b2−4ac2a=−(q2−p2)±(q2−p2)2−4(p2)(−q2)2p2=p2−q2±q4+p4−2q2p2+4p2q22p2=p2−q2±q4+p4+2p2q22p2=p2−q2±(q2+p2)22p2=(p2−q2)±(p2+q2)2p2=(p2−q2)+(p2+q2)2p2 or (p2−q2)−(p2+q2)2p2=2p22p2 or −2q22p2=1 or −q2p2.\Rightarrow x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[1em] = \dfrac{-(q^2 - p^2) \pm \sqrt{(q^2 - p^2)^2 - 4(p^2)(-q^2)}}{2p^2} \\[1em] = \dfrac{p^2 - q^2 \pm \sqrt{q^4 + p^4 - 2q^2p^2 + 4p^2q^2}}{2p^2} \\[1em] = \dfrac{p^2 - q^2 \pm \sqrt{q^4 + p^4 + 2p^2q^2}}{2p^2} \\[1em] = \dfrac{p^2 - q^2 \pm \sqrt{(q^2 + p^2)^2}}{2p^2} \\[1em] = \dfrac{(p^2 - q^2) \pm (p^2 + q^2)}{2p^2} \\[1em] = \dfrac{(p^2 - q^2) + (p^2 + q^2)}{2p^2} \text{ or } \dfrac{(p^2 - q^2) - (p^2 + q^2)}{2p^2} \\[1em] = \dfrac{2p^2}{2p^2} \text{ or } \dfrac{-2q^2}{2p^2} \\[1em] = 1 \text{ or } -\dfrac{q^2}{p^2}.

Hence, x = 1, −q2p2.-\dfrac{q^2}{p^2}.

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