If the sum of two of the roots of {x^3} + p{x | vedclass

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

Question

(B) Let the roots of the cubic equation ${x^3} + p{x^2} + qx + r = 0$ be $\alpha, \beta, \gamma$.Given that the sum of two roots is zero,let $\alpha +

Vedclass · Accepted Answer

$r$

Vedclass · Answer

(B) Let the roots of the cubic equation ${x^3} + p{x^2} + qx + r = 0$ be $\alpha, \beta, \gamma$.Given that the sum of two roots is zero,let $\alpha + \beta = 0$.From the relation between roots and coefficients,the sum of the roots is $\alpha + \beta + \gamma = -p$.Substituting $\alpha + \beta = 0$,we get $0 + \gamma = -p$,so $\gamma = -p$.Since $\gamma$ is a root of the equation,it must satisfy the equation: $(-p)^3 + p(-p)^2 + q(-p) + r = 0$.$-p^3 + p^3 - pq + r = 0$.$-pq + r = 0$.Therefore,$pq = r$.

If the sum of two of the roots of ${x^3} + p{x^2} + qx + r = 0$ is zero,then $pq =$

Similar Questions

If $\alpha$ and $\beta$ are the roots of the equation $x^2 + px + q = 0$,then what is the equation whose roots are $q/\alpha$ and $q/\beta$?

The sum of the cubes of all the roots of the equation $x^{4}-3x^{3}-2x^{2}+3x+1=10$ is

Let $a, b, c$ be the roots of the equation $x^3 + 8x + 1 = 0$. Then the value of $\frac{bc}{(8b + 1)(8c + 1)} + \frac{ac}{(8a + 1)(8c + 1)} + \frac{ab}{(8a + 1)(8b + 1)}$ is equal to

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+ax^2+bx+c=0$,then $\alpha^{-1}+\beta^{-1}+\gamma^{-1}$ is equal to

The condition that the roots of $x^3-b x^2+c x-d=0$ are in geometric progression is

Vedclass Test Series

Exam Paper Generator

Online Exam Module