The roots of the cubic equation 3x^3+4x^2-5x- | vedclass

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(C) Given equation: $3x^3+4x^2-5x-2=0$ $(i)$.Let the roots of $(i)$ be $\alpha, \beta, \gamma$.The roots are diminished by $h$,so the new roots are $t

Vedclass · Accepted Answer

$\frac{13}{9}, \frac{-14}{9}, \frac{1}{9}$

Vedclass · Answer

(C) Given equation: $3x^3+4x^2-5x-2=0$ $(i)$.Let the roots of $(i)$ be $\alpha, \beta, \gamma$.The roots are diminished by $h$,so the new roots are $t = x - h$,which implies $x = t + h$.Substituting $x = t + h$ into $(i)$:$3(t+h)^3 + 4(t+h)^2 - 5(t+h) - 2 = 0$.Expanding the terms:$3(t^3 + 3t^2h + 3th^2 + h^3) + 4(t^2 + 2th + h^2) - 5(t+h) - 2 = 0$.$3t^3 + (9h + 4)t^2 + (9h^2 + 8h - 5)t + (3h^3 + 4h^2 - 5h - 2) = 0$.Since the $x^2$ (or $t^2$) term is absent,its coefficient must be zero:$9h + 4 = 0 \Rightarrow h = -\frac{4}{9}$.Now,find the roots of the original equation $3x^3+4x^2-5x-2=0$.By inspection,$x=1$ is a root $(3+4-5-2=0)$.Dividing by $(x-1)$,we get $(x-1)(3x^2+7x+2) = 0$.$(x-1)(3x+1)(x+2) = 0$.The roots are $x_1 = 1, x_2 = -\frac{1}{3}, x_3 = -2$.The new roots are $x_i - h = x_i - (-\frac{4}{9}) = x_i + \frac{4}{9}$.New roots: $1 + \frac{4}{9} = \frac{13}{9}$,$-\frac{1}{3} + \frac{4}{9} = \frac{1}{9}$,and $-2 + \frac{4}{9} = -\frac{14}{9}$.

The roots of the cubic equation $3x^3+4x^2-5x-2=0$ are diminished by $h$,and a cubic equation with these diminished roots is formed. If the transformed equation does not contain the $x^2$ term,then the roots of the transformed equation are

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