If the roots of the equation x^3+3px^2+3qx-8= | vedclass

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

Question

(D) Let the roots of the cubic equation $x^3+3px^2+3qx-8=0$ be $\frac{a}{r}, a, ar$. From the relation between roots and coefficients,the product of t

Vedclass · Accepted Answer

-$8$

Vedclass · Answer

(D) Let the roots of the cubic equation $x^3+3px^2+3qx-8=0$ be $\frac{a}{r}, a, ar$. From the relation between roots and coefficients,the product of the roots is $\frac{a}{r} 	imes a 	imes ar = -(-8) = 8$. Thus,$a^3 = 8$,which implies $a = 2$. Since $a=2$ is a root,it must satisfy the equation: $(2)^3 + 3p(2)^2 + 3q(2) - 8 = 0$. $8 + 12p + 6q - 8 = 0$,which simplifies to $12p + 6q = 0$,or $q = -2p$. Substituting $q = -2p$ into the expression $\frac{q^3}{p^3}$,we get $\frac{(-2p)^3}{p^3} = \frac{-8p^3}{p^3} = -8$.

If the roots of the equation $x^3+3px^2+3qx-8=0$ are in a geometric progression,then $\frac{q^3}{p^3}=$

Similar Questions

If $\alpha, \beta$ are roots of the equation $x^{2}+5 \sqrt{2} x+10=0$,$\alpha > \beta$ and $P_{n}=\alpha^{n}-\beta^{n}$ for each positive integer $n$,then the value of $\left(\frac{P_{17} P_{20}+5 \sqrt{2} P_{17} P_{19}}{P_{18} P_{19}+5 \sqrt{2} P_{18}^{2}}\right)$ is equal to $....$

If $\alpha$ and $\beta$ are the roots of the equation $2x(2x+1)=1$,then $\beta$ is equal to

If the ratio of the roots of $x^2 + bx + c = 0$ and $x^2 + qx + r = 0$ is the same,then

If the sum of two roots of the equation $x^4+2x^3-7x^2-8x+12=0$ is zero,then the sum of the squares of the other two roots is

If $x^2 + px + 1$ is a factor of $ax^3 + bx + c$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module