Let alpha and beta be the roots of x^{2}-3x+p | vedclass

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(C) Let the roots be $\alpha, \alpha r, \alpha r^2, \alpha r^3$ in geometric progression.From the first equation $x^2 - 3x + p = 0$,the sum of roots i

Vedclass · Accepted Answer

$9:7$

Vedclass · Answer

(C) Let the roots be $\alpha, \alpha r, \alpha r^2, \alpha r^3$ in geometric progression.From the first equation $x^2 - 3x + p = 0$,the sum of roots is $\alpha + \alpha r = 3 \implies \alpha(1+r) = 3$ (Equation $1$).The product of roots is $p = \alpha^2 r$.From the second equation $x^2 - 6x + q = 0$,the sum of roots is $\alpha r^2 + \alpha r^3 = 6 \implies \alpha r^2(1+r) = 6$ (Equation $2$).The product of roots is $q = \alpha^2 r^5$.Dividing Equation $2$ by Equation $1$: $\frac{\alpha r^2(1+r)}{\alpha(1+r)} = \frac{6}{3} \implies r^2 = 2$.Now,calculate $p$ and $q$ in terms of $\alpha$ and $r$: $p = \alpha^2 r$ and $q = \alpha^2 r^5 = p r^4$.Since $r^2 = 2$,then $r^4 = 4$.Thus,$q = p(4) = 4p$.The ratio $\frac{2q+p}{2q-p} = \frac{2(4p)+p}{2(4p)-p} = \frac{8p+p}{8p-p} = \frac{9p}{7p} = \frac{9}{7}$.

Let $\alpha$ and $\beta$ be the roots of $x^{2}-3x+p=0$ and $\gamma$ and $\delta$ be the roots of $x^{2}-6x+q=0$. If $\alpha, \beta, \gamma, \delta$ form a geometric progression,then the ratio $(2q+p):(2q-p)$ is:

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