Let alpha neq 1 be a real root of the equatio | vedclass

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

Question

(C) Given the equation $x^3-a x^2+a x-1=0$. We can factorize the equation as follows: $(x^3-1) - a x(x-1) = 0$ $(x-1)(x^2+x+1) - a x(x-1) = 0$ $(x-1)(

Vedclass · Accepted Answer

$\frac{1}{\alpha}$

Vedclass · Answer

(C) Given the equation $x^3-a x^2+a x-1=0$. We can factorize the equation as follows: $(x^3-1) - a x(x-1) = 0$ $(x-1)(x^2+x+1) - a x(x-1) = 0$ $(x-1)(x^2+x+1-a x) = 0$ $(x-1)(x^2+(1-a)x+1) = 0$. Since $\alpha 
eq 1$ is a root,$\alpha$ must satisfy the quadratic equation $x^2+(1-a)x+1=0$. Thus,$\alpha^2+(1-a)\alpha+1=0$. Dividing by $\alpha$ (since $\alpha 
eq 0$ as $0$ is not a root),we get $\alpha + (1-a) + \frac{1}{\alpha} = 0$,which implies $\frac{1}{\alpha} = a - 1 - \alpha$. Alternatively,note that if $x$ is a root,then $\frac{1}{x}$ is also a root because the equation is reciprocal. Substituting $x = \frac{1}{y}$ into the original equation: $(\frac{1}{y})^3 - a(\frac{1}{y})^2 + a(\frac{1}{y}) - 1 = 0$ $1 - ay + ay^2 - y^3 = 0$ $y^3 - ay^2 + ay - 1 = 0$. This is the same equation as the original. Therefore,if $\alpha$ is a root,then $\frac{1}{\alpha}$ is also a root.

Let $\alpha \neq 1$ be a real root of the equation $x^3-a x^2+a x-1=0$,where $a \neq -1$ is a real number. Then,a root of this equation,among the following,is

Similar Questions

Sum of the solutions of the equation $[x^2] - 2x + 1 = 0$ is (where $[.]$ denotes the greatest integer function).

If $\alpha$ is a repeated root of multiplicity $2$ of the equation $18x^3-33x^2+20x-4=0$,then

If $a > 0,$ then $\sqrt {a + \sqrt {a + \sqrt {a + ....\infty } } } $ is

What is the number of real solutions for the equation $x^2 - 3|x| + 2 = 0$?

If $a < b < c < d$,then the roots of the equation $(x - a)(x - c) + 2(x - b)(x - d) = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module