If the roots of the equation 8 x^3+6 p x^2+3 | vedclass

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(A) Let the roots of the equation $8 x^3+6 p x^2+3 q x-27=0$ be $\frac{a}{r}, a, ar$. From the relation between roots and coefficients: Sum of roots:

Vedclass · Accepted Answer

-$3$

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(A) Let the roots of the equation $8 x^3+6 p x^2+3 q x-27=0$ be $\frac{a}{r}, a, ar$. From the relation between roots and coefficients: Sum of roots: $\frac{a}{r}+a+ar = -\frac{6p}{8} = -\frac{3p}{4} \Rightarrow a(\frac{1}{r}+1+r) = -\frac{3p}{4} \dots (i)$ Sum of roots taken two at a time: $(\frac{a}{r} \cdot a) + (a \cdot ar) + (\frac{a}{r} \cdot ar) = \frac{3q}{8}$ $\Rightarrow a^2(\frac{1}{r}+r+1) = \frac{3q}{8} \dots (ii)$ Product of roots: $\frac{a}{r} \cdot a \cdot ar = -(\frac{-27}{8}) = \frac{27}{8}$ $\Rightarrow a^3 = \frac{27}{8}$ $\Rightarrow a = \frac{3}{2} \dots (iii)$ Dividing (ii) by $(i)$: $\frac{a^2(\frac{1}{r}+r+1)}{a(\frac{1}{r}+r+1)} = \frac{3q/8}{-3p/4}$ $\Rightarrow a = \frac{3q}{8} \cdot (-\frac{4}{3p}) = -\frac{q}{2p} \dots (iv)$ Equating (iii) and (iv): $\frac{3}{2} = -\frac{q}{2p} \Rightarrow q = -3p$. Now,substitute $q = -3p$ into the expression $q^2+9p^2+6pq+\frac{q}{p}$: $(-3p)^2 + 9p^2 + 6p(-3p) + \frac{-3p}{p} = 9p^2 + 9p^2 - 18p^2 - 3 = 18p^2 - 18p^2 - 3 = -3$.

If the roots of the equation $8 x^3+6 p x^2+3 q x-27=0$ are in a geometric progression,then $q^2+9 p^2+6 p q+q/p=$

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