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

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(B) Given the equation $x^3 + 5x^2 - 7x - 1 = 0$ with roots $\alpha, \beta, \gamma$.From Vieta's formulas,the product of the roots is $\alpha\beta\gam

Vedclass · Accepted Answer

$x^3 + 7x^2 - 5x - 1 = 0$

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(B) Given the equation $x^3 + 5x^2 - 7x - 1 = 0$ with roots $\alpha, \beta, \gamma$.From Vieta's formulas,the product of the roots is $\alpha\beta\gamma = -(-1)/1 = 1$.We want to find the equation whose roots are $\alpha\beta, \beta\gamma, \gamma\alpha$.Since $\alpha\beta\gamma = 1$,we can write the roots as:$\alpha\beta = \frac{\alpha\beta\gamma}{\gamma} = \frac{1}{\gamma}$,$\beta\gamma = \frac{\alpha\beta\gamma}{\alpha} = \frac{1}{\alpha}$,$\gamma\alpha = \frac{\alpha\beta\gamma}{\beta} = \frac{1}{\beta}$.Thus,the roots are $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$.To find the equation with roots $\frac{1}{x}$,we substitute $x$ with $\frac{1}{x}$ in the original equation:$(\frac{1}{x})^3 + 5(\frac{1}{x})^2 - 7(\frac{1}{x}) - 1 = 0$.Multiplying by $x^3$,we get:$1 + 5x - 7x^2 - x^3 = 0$,which simplifies to $x^3 + 7x^2 - 5x - 1 = 0$.

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3 + 5x^2 - 7x - 1 = 0$,then find the equation whose roots are $\alpha\beta, \beta\gamma, \gamma\alpha$.

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