Question

The root of the equation, $(x^2+1{)}^2=x(3x^2+4x+3),$ are given by

• A. $2-\sqrt{3}$
• B. $\frac{(-1+i\sqrt{3})}{2},i=\sqrt{-1}$
• C. $2+\sqrt{3}$
• D. $\frac{(-1-i\sqrt{3})}{2},i=\sqrt{-1}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A

$2-\sqrt{3}$

B

$\frac{(-1+i\sqrt{3})}{2},i=\sqrt{-1}$

C

$2+\sqrt{3}$

D

$\frac{(-1-i\sqrt{3})}{2},i=\sqrt{-1}$

Given equation is

$(x^2+1{)}^2=x(3x^2+4x+3)$

$\Rightarrow x^4-3x^3-2x^2-3x+1=0$

$\Rightarrow x^2(x^2-3x-2-\frac{3}{x}+\frac{1}{x^2})=0$

$\Rightarrow \because x\ne 0$

$\therefore x^2+\frac{1}{x^2}-3(x+\frac{1}{x})-2=0$

$\Rightarrow {(x+\frac{1}{x})}^2-3(x+\frac{1}{x})-4=0$

$\Rightarrow (x+\frac{1}{x}-4)(x+\frac{1}{x}+1)=0$

Or $(x^2-4x+1)(x^2+x+1)=0$

$\therefore x=2\pm \sqrt{3},\frac{-1\pm i\sqrt{3}}{2}$