Question

The range of values of a for which all the roots of the equation $(a-1){(1+x+x^2)}^2=(a+1)(1+x^2+x^4)$ imaginary is

• A. $(-\infty ,-2]$
• B. $(2,+\infty )$
• C. $(-2,2)$
• D. None of these

Answer 1

Answer: (A)

$(-\infty ,-2]$

$\frac{a-1}{a+1}=\frac{1+x^2+x^4}{{(1+x+x^2)}^2}$ by evaluation from the question

$=\frac{x^4+x^2+1}{{(1+x+x^2)}^2}$

$=\frac{{\left(x^2\right)}^2+2x^2+1-x^2}{{(1+x+x^2)}^2}$ arranging the numerator in the form ${(a+b)}^2$

$=\frac{{(x^2+1)}^2-x^2}{{(1+x+x^2)}^2}$ where the numerator is of the form $a^2-b^2$

$=\frac{(x^2+x+1)(x^2-x+1)}{{(1+x+x^2)}^2}$ where $a^2-b^2=(a-b)(a+b)$

$=\frac{x^2-x+1}{x^2+x+1}$ by cancelling the like terms in the numerator and denominator.

By using componendo-dividendo theorem, we have

$\Rightarrow \frac{a-1-a-1}{a-1+a+1}=\frac{x^2+1-x-x^2-x-1}{x^2+1-x+x^2+x+1}$

$\Rightarrow \frac{-2}{2a}=\frac{-2x}{2(x^2+1)}$

$\Rightarrow \frac{1}{a}=\frac{x}{(x^2+1)}$ by simplification

$\Rightarrow a=\frac{(x^2+1)}{x}$ by taking the reciprocal

$\Rightarrow a=x+\frac{1}{x}$ by seperating the terms.

To find the range we have

$a=x+\frac{1}{x}$

Range of function means set of all possible values of $a$

so we need to find all possible values if $a$

$\Rightarrow xa=x^2+1$

$\Rightarrow x^2-xa+1=0$

Discriminant of this quadratic function must be $<0$

as $x$ can only assume imaginary roots /values for all possible $a$

$D<0$ ( for a quadratic equation of for $a{x}^2+bx+c=0$ is given by

$b^2-4ac<0$)

which gives us

$a^2-4\left(1\right)\left(1\right)<0$

$a^2-4<0$

so $a$ belongs to $(-\infty ,-2]$

so Range of the function is $(-\infty ,-2]$