If the roots of x^4+x^3-4x^2+x+1=0 are dimini | vedclass

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(A) Let the roots of the equation $x^4+x^3-4x^2+x+1=0$ be diminished by $h$. Substituting $x$ with $x+h$,the equation becomes $(x+h)^4+(x+h)^3-4(x+h)^

Vedclass · Accepted Answer

$35$

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(A) Let the roots of the equation $x^4+x^3-4x^2+x+1=0$ be diminished by $h$. Substituting $x$ with $x+h$,the equation becomes $(x+h)^4+(x+h)^3-4(x+h)^2+(x+h)+1=0$. Expanding the terms,the coefficient of $x^2$ is given by $6h^2+3h-4$. Since the $x^2$ term is absent,we set $6h^2+3h-4=0$. The roots of this quadratic equation are $\alpha$ and $\beta$. Using the relation for the difference of roots,$(\alpha-\beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta$. From the quadratic equation $6h^2+3h-4=0$,we have $\alpha+\beta = -\frac{3}{6} = -\frac{1}{2}$ and $\alpha\beta = -\frac{4}{6} = -\frac{2}{3}$. Thus,$(\alpha-\beta)^2 = (-\frac{1}{2})^2 - 4(-\frac{2}{3}) = \frac{1}{4} + \frac{8}{3} = \frac{3+32}{12} = \frac{35}{12}$. Therefore,$12(\alpha-\beta)^2 = 12 	imes \frac{35}{12} = 35$.

If the roots of $x^4+x^3-4x^2+x+1=0$ are diminished by $\alpha$ or $\beta$,then the equation with the diminished roots does not contain the $x^2$ term. Then $12(\alpha-\beta)^2=$

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