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

The roots $\alpha, \beta$ of the equation $x^2-6(k-1)x+4(k-2)=0$ are equal in magnitude but opposite in sign. If $\alpha > \beta$,then the product of the roots of the equation $2x^2-\alpha x+6\beta(\alpha+1)=0$ is

If $x = 2 + \sqrt{3}$ and $xy = 1$,then $\frac{x}{\sqrt{2} + \sqrt{x}} + \frac{y}{\sqrt{2} - \sqrt{y}} = $

The number of distinct real roots of the equation $|x||x+2|-5|x+1|-1=0$ is:

If $x$ is a solution of the equation $\sqrt{2x + 1} - \sqrt{2x - 1} = 1$ for $x \ge \frac{1}{2}$,then $\sqrt{4x^2 - 1}$ is equal to:

The solution of the equation $x = \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots}}}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module