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Mathematics

The roots of the equation px2 + qx + r = 0, where p ≠ 0, are given by:

  1. x=−p±q2−4pr2px = \dfrac{-p \pm \sqrt{q^{2} - 4pr}}{2p}

  2. x=−q±q2−2pr4px = \dfrac{-q \pm \sqrt{q^{2} - 2pr}}{4p}

  3. x=−q±q2−4pr2px = \dfrac{-q \pm \sqrt{q^{2} - 4pr}}{2p}

  4. x=−q±q2−4pr2qx = \dfrac{-q \pm \sqrt{q^{2} - 4pr}}{2q}

Quadratic Equations

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Answer

Given,

⇒ px2 + qx + r = 0

Comparing equation px2 + qx + r = 0 with ax2 + bx + c = 0, we get :

a = p, b = q and c = r.

By formula,

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

Substituting values we get :

x = −q±q2−4pr2p\dfrac{-q \pm \sqrt{q^2 - 4pr}}{2p}.

Hence, option 3 is the correct option.

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