If $\alpha, \beta, \gamma, \delta$ are in GP where $\alpha, \beta$ are roots of the equation $ax^2 + 2bx + c = 0$ and $\gamma, \delta$ are roots of the equation $px^2 + 2qx + r = 0,$ then
- $\displaystyle{\frac{ac}{b^2} = \frac{pr}{q^2}}$
- $\displaystyle{\frac{ac}{b} = \frac{pr}{q}}$
- $\displaystyle{\frac{ab}{c^2} = \frac{pq}{r^2}}$
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None of these
For ax^2 + 2bx + c = 0, roots alpha, beta are in GP, so alpha*beta = c/a. For px^2 + 2qx + r = 0, roots gamma, delta are in GP, so gamma*delta = r/p. Since alpha, beta, gamma, delta are in GP, beta/alpha = delta/gamma = k. Also beta = alpha*k, gamma = beta*k = alpha*k^2, delta = gamma*k = alpha*k^3. Then c/a = alpha^2 * k and r/p = alpha^2 * k^5. This leads to the ratio ac/b^2 = pr/q^2.
Since alpha and beta are the roots of ax squared plus 2bx plus c equals 0, their product is alpha beta equals c divided by a. Because gamma and delta are the roots of px squared plus 2qx plus r equals 0, their product is gamma delta equals r divided by p. All four variables are in a geometric progression, so the square of any term equals the product of its equidistant neighbors. Specifically, beta squared equals alpha gamma and gamma squared equals beta delta. Multiplying these gives beta squared gamma squared equals alpha beta gamma delta, which rearranges to beta gamma equals alpha beta gamma delta divided by beta gamma. Substituting the sum and product relationships yields ac divided by b squared equals pr divided by q squared. This matches the required relation.