If one root of the equation x^2 + px + q = 0 | vedclass

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

Question

(D) Let the roots of the given equation $x^2 + px + q = 0$ be $\alpha$ and $\alpha^2$.From the relationship between roots and coefficients:Product of

Vedclass · Accepted Answer

$p^3 + q^2 + q(1 - 3p) = 0$

Vedclass · Answer

(D) Let the roots of the given equation $x^2 + px + q = 0$ be $\alpha$ and $\alpha^2$.From the relationship between roots and coefficients:Product of roots: $\alpha \cdot \alpha^2 = \alpha^3 = q$Sum of roots: $\alpha + \alpha^2 = -p$Now,cube both sides of the sum equation: $(\alpha + \alpha^2)^3 = (-p)^3$Using the identity $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$:$\alpha^3 + (\alpha^2)^3 + 3\alpha \cdot \alpha^2(\alpha + \alpha^2) = -p^3$Substitute $\alpha^3 = q$ and $\alpha + \alpha^2 = -p$ into the equation:$q + q^2 + 3q(-p) = -p^3$Rearranging the terms,we get:$p^3 + q^2 + q - 3pq = 0$$p^3 + q^2 + q(1 - 3p) = 0$

If one root of the equation $x^2 + px + q = 0$ is the square of the other,then

Similar Questions

If $\alpha_1, \alpha_2$ and $\beta_1, \beta_2$ are the roots of the equations $ax^2 + bx + c = 0$ and $px^2 + qx + r = 0$ respectively,and the system of equations $\alpha_1 y + \alpha_2 z = 0$ and $\beta_1 y + \beta_2 z = 0$ has a non-zero solution,then:

The complete solution set of the inequality $(\sec^{-1}x - 4)(\sec^{-1}x - 1)(\sec^{-1}x - 2) \ge 0$ is

The roots of the equation $x^2 - \sqrt{13}x + 1 = 0$ are:

If the roots of the given equation $2x^2 + 3(\lambda - 2)x + \lambda + 4 = 0$ are equal in magnitude but opposite in sign,then $\lambda = $

If the roots of the given equation $(\cos p - 1)x^2 + (\cos p)x + \sin p = 0$ are real,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module