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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$35$

Vedclass · Answer

(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 $12(\alpha-\beta)^2=$

Similar Questions

Let $\theta$,$0 < \theta < \pi / 2$,be an angle such that the equation $x^2 + 4x \cos \theta + \cot \theta = 0$ has equal roots for $x$. Then $\theta$ in radians is

The quadratic equation whose one root is $\frac{1}{2 + \sqrt{5}}$ will be

Solve the equation,$3^{x^2-x}=25-4^{x^2-x}$

The number of all possible positive integral values of $\alpha$ for which the roots of the quadratic equation $6x^2 - 11x + \alpha = 0$ are rational numbers is:

If the equation $x^2 + \lambda x + \mu = 0$ has equal roots and one root of the equation $x^2 + \lambda x - 12 = 0$ is $2$,then $(\lambda, \mu) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module