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

(C) Given that $\alpha, \beta, \gamma$ are the roots of the equation $x^3+0x^2+4x+1=0$. From Vieta's formulas,we have: $\alpha+\beta+\gamma = 0$ $\alp

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Given that $\alpha, \beta, \gamma$ are the roots of the equation $x^3+0x^2+4x+1=0$. From Vieta's formulas,we have: $\alpha+\beta+\gamma = 0$ $\alpha\beta+\beta\gamma+\gamma\alpha = 4$ $\alpha\beta\gamma = -1$ Since $\alpha+\beta+\gamma = 0$,we can write: $\alpha+\beta = -\gamma$ $\beta+\gamma = -\alpha$ $\gamma+\alpha = -\beta$ Substituting these into the expression: $(\alpha+\beta)^{-1}+(\beta+\gamma)^{-1}+(\gamma+\alpha)^{-1} = \frac{1}{-\gamma} + \frac{1}{-\alpha} + \frac{1}{-\beta}$ $= -\left(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}\right) = -\left(\frac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma}\right)$ $= -\left(\frac{4}{-1}\right) = 4$

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+4x+1=0$,then $(\alpha+\beta)^{-1}+(\beta+\gamma)^{-1}+(\gamma+\alpha)^{-1}$ is equal to

Similar Questions

If $\frac{x-4}{x^2-5x-2k} = \frac{2}{x-2} - \frac{1}{x+k}$,then $k$ is equal to

For the equation $ax^2 + bx + c = 0$,if $\alpha$ and $\beta$ are its roots,then the equation whose roots are $\alpha + \frac{1}{\beta}$ and $\beta + \frac{1}{\alpha}$ is:

If the harmonic mean between the roots of $(5+\sqrt{2}) x^2-b x+(8+2 \sqrt{5})=0$ is $4$,then the value of $b$ is

If the ratio of the roots of $ax^2 + 2bx + c = 0$ is the same as the ratio of the roots of $px^2 + 2qx + r = 0$,then

If the equation having the roots as the values obtained by diminishing each root of the equation $x^3-3x^2+2x-1=0$ by $K$ is $x^3-x-1=0$,then $K=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module