The roots of the equation x^4+x^3-4x^2+x+1=0 | vedclass

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(A) The given equation is $x^4+x^3-4x^2+x+1=0$. Dividing by $x^2$,we get $(x^2 + \frac{1}{x^2}) + (x + \frac{1}{x}) - 4 = 0$. Let $y = x + \frac{1}{x}

Vedclass · Accepted Answer

$35$

Vedclass · Answer

(A) The given equation is $x^4+x^3-4x^2+x+1=0$. Dividing by $x^2$,we get $(x^2 + \frac{1}{x^2}) + (x + \frac{1}{x}) - 4 = 0$. Let $y = x + \frac{1}{x}$,then $y^2 - 2 + y - 4 = 0 \Rightarrow y^2 + y - 6 = 0$. $(y+3)(y-2) = 0$,so $y = 2$ or $y = -3$. If $x + \frac{1}{x} = 2$,then $x=1, 1$. If $x + \frac{1}{x} = -3$,then $x^2 + 3x + 1 = 0$,so $x = \frac{-3 \pm \sqrt{5}}{2}$. Let the roots be $r_1, r_2, r_3, r_4$. The sum of roots taken two at a time is $\sum r_i r_j = -4$. When roots are diminished by $h$,the new roots are $r_i - h$. The new coefficient of $x^2$ is $\sum (r_i - h)(r_j - h) = 0$. This expands to $\sum r_i r_j - 3h \sum r_i + 6h^2 = 0$. From the original equation,$\sum r_i = -1$ and $\sum r_i r_j = -4$. Substituting these,$-4 - 3h(-1) + 6h^2 = 0 \Rightarrow 6h^2 + 3h - 4 = 0$. The roots of this quadratic in $h$ are $\alpha$ and $\beta$. Thus,$\alpha + \beta = -\frac{3}{6} = -\frac{1}{2}$ and $\alpha \beta = -\frac{4}{6} = -\frac{2}{3}$. Then $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta = (-\frac{1}{2})^2 - 4(-\frac{2}{3}) = \frac{1}{4} + \frac{8}{3} = \frac{3+32}{12} = \frac{35}{12}$. Therefore,$12(\alpha - \beta)^2 = 12 	imes \frac{35}{12} = 35$.

The roots of the equation $x^4+x^3-4x^2+x+1=0$ are diminished by $h$ so that the transformed equation does not contain the $x^2$ term. If the values of such $h$ are $\alpha$ and $\beta$,then $12(\alpha-\beta)^2=$

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