If alpha, beta, gamma, delta are in geometric | vedclass

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Question

(A) Since $\alpha, \beta, \gamma, \delta$ are in geometric progression,let the common ratio be $k$. Then $\beta = \alpha k$,$\gamma = \alpha k^2$,and

Vedclass · Accepted Answer

$\frac{ac}{b^2} = \frac{pr}{q^2}$

Vedclass · Answer

(A) Since $\alpha, \beta, \gamma, \delta$ are in geometric progression,let the common ratio be $k$. Then $\beta = \alpha k$,$\gamma = \alpha k^2$,and $\delta = \alpha k^3$.For the equation $ax^2 + 2bx + c = 0$,the sum of roots $\alpha + \beta = -\frac{2b}{a}$ and product $\alpha \beta = \frac{c}{a}$.For the equation $px^2 + 2qx + r = 0$,the sum of roots $\gamma + \delta = -\frac{2q}{p}$ and product $\gamma \delta = \frac{r}{p}$.Since $\alpha, \beta, \gamma, \delta$ are in $GP$,$\frac{\beta}{\alpha} = \frac{\delta}{\gamma} = k$.Also,$\frac{\alpha + \beta}{\alpha - \beta} = \frac{1+k}{1-k}$ and $\frac{\gamma + \delta}{\gamma - \delta} = \frac{\alpha k^2 + \alpha k^3}{\alpha k^2 - \alpha k^3} = \frac{1+k}{1-k}$.Thus,$\frac{(\alpha + \beta)^2}{\alpha \beta} = \frac{(\alpha + \alpha k)^2}{\alpha^2 k} = \frac{\alpha^2(1+k)^2}{\alpha^2 k} = \frac{(1+k)^2}{k}$.Similarly,$\frac{(\gamma + \delta)^2}{\gamma \delta} = \frac{(\alpha k^2 + \alpha k^3)^2}{\alpha^2 k^5} = \frac{\alpha^2 k^4(1+k)^2}{\alpha^2 k^5} = \frac{(1+k)^2}{k}$.Therefore,$\frac{(\alpha + \beta)^2}{\alpha \beta} = \frac{(\gamma + \delta)^2}{\gamma \delta}$.Substituting the coefficients: $\frac{(-2b/a)^2}{c/a} = \frac{(-2q/p)^2}{r/p}$ $\Rightarrow \frac{4b^2/a^2}{c/a} = \frac{4q^2/p^2}{r/p}$ $\Rightarrow \frac{4b^2}{ac} = \frac{4q^2}{pr}$.This simplifies to $\frac{b^2}{ac} = \frac{q^2}{pr}$,which is equivalent to $\frac{ac}{b^2} = \frac{pr}{q^2}$.

If $\alpha, \beta, \gamma, \delta$ are in geometric progression,where $\alpha, \beta$ are the roots of the equation $ax^2 + 2bx + c = 0$ and $\gamma, \delta$ are the roots of the equation $px^2 + 2qx + r = 0$,then:

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