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If the roots of $x^4+x^3-4 x^2+x+1=0$ are diminished by $\alpha$ or $\beta$, then the equation with the diminished roots does not contain $x^2$ term. Then $12(\alpha-\beta)^2=$
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Verified Answer
The correct answer is:
$35$
Let the roots of equation
$x^4+x^3-4 x^2+x+1=0$ are diminished by $h$, so the equation becomes
$(x+h)^4+(x+h)^3-4(x+h)^2+(x+h)+1=0$
Now the coefficient of $x^2$ is $6 h^2+3 h-4$ must be zero
$\therefore \quad 6 h^2+3 h-4=0$ having roots $\alpha$ and $\beta$
So, $(\alpha-\beta)^2=(\alpha+\beta)^2-4 \alpha \beta$
$\begin{aligned} & =\left(\frac{-3}{6}\right)^2-4\left(-\frac{4}{6}\right) \\ & =\frac{1}{4}+\frac{8}{3}=\frac{3+32}{12}=\frac{35}{12}\end{aligned}$
Therefore $12(\alpha-\beta)^2=35$
$x^4+x^3-4 x^2+x+1=0$ are diminished by $h$, so the equation becomes
$(x+h)^4+(x+h)^3-4(x+h)^2+(x+h)+1=0$
Now the coefficient of $x^2$ is $6 h^2+3 h-4$ must be zero
$\therefore \quad 6 h^2+3 h-4=0$ having roots $\alpha$ and $\beta$
So, $(\alpha-\beta)^2=(\alpha+\beta)^2-4 \alpha \beta$
$\begin{aligned} & =\left(\frac{-3}{6}\right)^2-4\left(-\frac{4}{6}\right) \\ & =\frac{1}{4}+\frac{8}{3}=\frac{3+32}{12}=\frac{35}{12}\end{aligned}$
Therefore $12(\alpha-\beta)^2=35$
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