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(B) Let the roots of the polynomial $f(x) = x^3 - 9x^2 + 26x - 24$ be $\alpha, \beta, \gamma$. To find the equation whose roots are the reciprocals $\

Vedclass · Accepted Answer

$24x^3 - 26x^2 + 9x - 1$

Vedclass · Answer

(B) Let the roots of the polynomial $f(x) = x^3 - 9x^2 + 26x - 24$ be $\alpha, \beta, \gamma$. To find the equation whose roots are the reciprocals $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$,we substitute $x$ with $\frac{1}{x}$ in the original equation $f(x) = 0$. $\left(\frac{1}{x}\right)^3 - 9\left(\frac{1}{x}\right)^2 + 26\left(\frac{1}{x}\right) - 24 = 0$. Multiplying the entire equation by $-x^3$,we get: $-1 + 9x - 26x^2 + 24x^3 = 0$. Rearranging the terms,we obtain $24x^3 - 26x^2 + 9x - 1 = 0$. Thus,the correct option is $B$.

If $\alpha, \beta, \gamma$ are the roots of $f(x) = x^3 - 9x^2 + 26x - 24$,then $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$ are the roots of which equation?

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