The condition that x^3 - p x^2 + q x - r = 0 | vedclass

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

Question

(A) Given the cubic equation $x^3 - p x^2 + q x - r = 0$ ... $(i)$.Let the roots be $\alpha, \beta, \gamma$.From Vieta's formulas:$\alpha + \beta + \g

Vedclass · Accepted Answer

$r = pq$

Vedclass · Answer

(A) Given the cubic equation $x^3 - p x^2 + q x - r = 0$ ... $(i)$.Let the roots be $\alpha, \beta, \gamma$.From Vieta's formulas:$\alpha + \beta + \gamma = p$ ... $(ii)$$\alpha \beta + \beta \gamma + \gamma \alpha = q$ ... $(iii)$$\alpha \beta \gamma = r$ ... $(iv)$Given that two roots are equal in magnitude but opposite in sign,let $\alpha = -\beta$.Substituting $\alpha = -\beta$ into $(ii)$:$-\beta + \beta + \gamma = p \implies \gamma = p$.Substituting $\gamma = p$ into $(iv)$:$\alpha \beta (p) = r \implies -\beta^2 p = r \implies \beta^2 = -\frac{r}{p}$.Substituting $\alpha = -\beta$ and $\gamma = p$ into $(iii)$:$-\beta^2 + \beta p - \beta p = q \implies -\beta^2 = q$.Equating the two expressions for $\beta^2$:$-\frac{r}{p} = -q \implies r = pq$.

The condition that $x^3 - p x^2 + q x - r = 0$ may have two of its roots equal to each other but of opposite sign is

Similar Questions

If $\alpha, \beta$ are the roots of $x^2 - px + q = 0$ and $\alpha', \beta'$ are the roots of $x^2 - p'x + q' = 0$,then the value of $(\alpha - \alpha')^2 + (\beta - \alpha')^2 + (\alpha - \beta')^2 + (\beta - \beta')^2$ is

If the roots of the equation $ax^2 + bx + c = 0$ are $\sin \alpha$ and $\cos \alpha$,then:

If $\alpha, \beta, \gamma$ are the real roots of the equation $x^3 - 3px^2 + 3qx - 1 = 0$,then find the centroid of the triangle whose vertices are $(\alpha, \frac{1}{\alpha}), (\beta, \frac{1}{\beta})$ and $(\gamma, \frac{1}{\gamma})$.

What is the sum of the squares of the roots of the equation $x^2 - 3x + 1 = 0$?

If $\alpha, \beta, \gamma, \delta$ are the roots of $x^4 - 100x^3 + 2x^2 + 4x + 10 = 0$,then $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\delta}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module