If 2, 1, 1 are roots of the equation x^3-4x^2 | vedclass

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(B) Let the given equation be $f(x) = x^3-4x^2+5x-2=0$,which has roots $x = 2, 1, 1$. The second equation is given by $f\left(x+\frac{1}{3}\right) = 0

Vedclass · Accepted Answer

$\frac{5}{3}, \frac{2}{3}, \frac{2}{3}$

Vedclass · Answer

(B) Let the given equation be $f(x) = x^3-4x^2+5x-2=0$,which has roots $x = 2, 1, 1$. The second equation is given by $f\left(x+\frac{1}{3}\right) = 0$. If $x_0$ is a root of $f(x) = 0$,then $x+\frac{1}{3} = x_0$ must hold for the new equation. Therefore,$x = x_0 - \frac{1}{3}$. Substituting the roots $x_0 = 2, 1, 1$ into this relation: $x_1 = 2 - \frac{1}{3} = \frac{5}{3}$ $x_2 = 1 - \frac{1}{3} = \frac{2}{3}$ $x_3 = 1 - \frac{1}{3} = \frac{2}{3}$ Thus,the roots of the new equation are $\frac{5}{3}, \frac{2}{3}, \frac{2}{3}$.

If $2, 1, 1$ are roots of the equation $x^3-4x^2+5x-2=0$,then find the roots of the equation $\left(x+\frac{1}{3}\right)^3-4\left(x+\frac{1}{3}\right)^2+5\left(x+\frac{1}{3}\right)-2=0$.

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