Question

The roots equation $x^4-3x^3+4x^2-3x+1=0$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B $1$

$x^4-3x^3+4x^2-3x+1=0$
This equation is resiprocal equation of first type as $a_{n-i}=a_i$
Dividing equation by $x^2$:
$x^2-3x+4-\frac{3}{x}+\frac{1}{x^2}=0$
$x^2+\frac{1}{x^2}-3x-\frac{3}{x}+4=0$
${(x+\frac{1}{x})}^2-2-3(x+\frac{1}{x})+4=0$
${(x+\frac{1}{x})}^2-3(x+\frac{1}{x})+2=0$
Let $x+\frac{1}{x}=y$
$y^2-3y+2=0$
$(y-1)(y-2)=0$
$y=1$ or $y=2$
For $y=1$:
$x+\frac{1}{x}=1$
$x^2-x+1=0$
$D=(-1{)}^2-4(1\left)\right(1)=-3<0$
Hence no real value of x exists for this case.
For $y=2$:
$x+\frac{1}{x}=2$
$x^2-2x+1=0$
$(x-1{)}^2=0$
$x=1$
Hence solution of the given equation is x=1.