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

Vedclass · Accepted Answer

$\frac{1}{\alpha}$

Vedclass · Answer

(C) Given equation: $x^3-ax^2+ax-1=0$. We can factorize the equation as: $(x^3-1) - ax(x-1) = 0$. $(x-1)(x^2+x+1) - ax(x-1) = 0$. $(x-1)(x^2+x+1-ax) = 0$. Since $\alpha 
eq 1$ is a root,$\alpha$ must satisfy $x^2+(1-a)x+1=0$. Thus,$\alpha^2+(1-a)\alpha+1=0$. If $x = \frac{1}{\alpha}$ is a root,then $(\frac{1}{\alpha})^3 - a(\frac{1}{\alpha})^2 + a(\frac{1}{\alpha}) - 1 = 0$. Multiplying by $\alpha^3$,we get $1 - a\alpha + a\alpha^2 - \alpha^3 = 0$,which is $-(x^3-ax^2+ax-1) = 0$ evaluated at $x=\alpha$. Since $\alpha$ is a root,$\frac{1}{\alpha}$ is also a root of the equation.

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

Similar Questions

An auditorium has a volume of $10^5 \ m^3$ and a total absorption area of $2 \times 10^4 \ m^2$. Its average absorption coefficient is $0.2$. The reverberation time of the auditorium in seconds is:

Which of the following is the most basic oxide?

Observe the following statements:
$I. f(x) = a x^{41} + b x^{-40} \Rightarrow \frac{f^{\prime \prime}(x)}{f(x)} = 1640 x^{-2}$
$II. \frac{d}{d x} \tan ^{-1}\left(\frac{2 x}{1-x^2}\right) = \frac{1}{1+x^2}$
Which of the following is correct?

Which selection method is used for vegetative propagation in plants?

Which of the following animals is not viviparous?

Vedclass Test Series

Exam Paper Generator

Online Exam Module