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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

-$3$

Vedclass · Answer

(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 $q^2+9 p^2+6 p q+q/p=$

Similar Questions

If the roots of the equation $x^5-40x^4-Px^3-Rx-S=0$ are in geometric progression and the sum of the reciprocals of the roots is $10$,then $|S|=$

If $x = \frac{4}{3} - \frac{4x}{9} + \frac{4x^2}{27} - \dots \infty$,then $x$ is equal to

If $A, B, C$ are the $p^{th}, q^{th},$ and $r^{th}$ terms of a $GP$ respectively,then $A^{q-r} \cdot B^{r-p} \cdot C^{p-q} =$

The sum of the first three terms of a $G.P.$ is $\frac{13}{12}$ and their product is $-1$. Find the common ratio and the terms.

If the $p^{\text{th}}$,$q^{\text{th}}$,and $r^{\text{th}}$ terms of a $G.P.$ are $a$,$b$,and $c$ respectively,prove that $a^{q-r} b^{r-p} c^{p-q} = 1$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module