If alpha, beta, gamma are the roots of the eq | vedclass

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

Question

(B) Given the equation $x^3 + qx - r = 0$,the roots are $\alpha, \beta, \gamma$. From Vieta's formulas: $\alpha + \beta + \gamma = 0$ $\alpha \beta +

Vedclass · Accepted Answer

$rx^3 - q(r + 1)x^2 - (r + 1)^3 = 0$

Vedclass · Answer

(B) Given the equation $x^3 + qx - r = 0$,the roots are $\alpha, \beta, \gamma$. From Vieta's formulas: $\alpha + \beta + \gamma = 0$ $\alpha \beta + \beta \gamma + \gamma \alpha = q$ $\alpha \beta \gamma = r$ We need to find the equation with roots $y_1 = \beta \gamma + \frac{1}{\alpha}$,$y_2 = \gamma \alpha + \frac{1}{\beta}$,and $y_3 = \alpha \beta + \frac{1}{\gamma}$. Since $\alpha \beta \gamma = r$,we have $\beta \gamma = \frac{r}{\alpha}$,$\gamma \alpha = \frac{r}{\beta}$,and $\alpha \beta = \frac{r}{\gamma}$. Thus,the roots are $y_1 = \frac{r}{\alpha} + \frac{1}{\alpha} = \frac{r+1}{\alpha}$,$y_2 = \frac{r+1}{\beta}$,and $y_3 = \frac{r+1}{\gamma}$. Let $y = \frac{r+1}{x}$,then $x = \frac{r+1}{y}$. Substitute $x$ into the original equation: $(\frac{r+1}{y})^3 + q(\frac{r+1}{y}) - r = 0$ $\frac{(r+1)^3}{y^3} + \frac{q(r+1)}{y} - r = 0$ Multiply by $y^3$: $(r+1)^3 + q(r+1)y^2 - ry^3 = 0$ $ry^3 - q(r+1)y^2 - (r+1)^3 = 0$. Replacing $y$ with $x$,we get $rx^3 - q(r+1)x^2 - (r+1)^3 = 0$.

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3 + qx - r = 0$,then find the equation whose roots are $\left( \beta \gamma + \frac{1}{\alpha} \right), \left( \gamma \alpha + \frac{1}{\beta} \right), \left( \alpha \beta + \frac{1}{\gamma} \right)$.

Similar Questions

Solve the given two equations and choose the correct option.
$I.$ $99 x^{2} + 149 x + 56 = 0$
$II.$ $156 y^{2} + 287 y + 132 = 0$

Let $p, q, r \in \mathbb{R}$ and $r > p > 0.$ If the quadratic equation $px^2 + qx + r = 0$ has two complex roots $\alpha$ and $\beta,$ then $|\alpha| + |\beta|$ is:

Product of real roots of the equation ${t^2}{x^2} + |x| + 9 = 0$.

Determine $p$ such that the equation $x^{2}+5px+16=0$ has no real roots.

If $\alpha$ and $\beta$ are the roots of the equation ${x^2} - a(x + 1) - b = 0$,then $(\alpha + 1)(\beta + 1) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module