Question

Solve the reciprocal equation $x^4-3x^3+4x^2-3x+1=0$

• A. $0$
• B. $1$
• C. $3$
• D. $-1$

Answer 1

The correct option is B $1$

We see that it is a reciprocal equation, so we divide it by $x^2$:

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

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

We will now substitute $x+\frac{1}{x}=u$

By squaring it, and solve further we get

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

We plug back into the equation to get

$\Rightarrow$$u^2-3u+2=0$

The solutions are $u_1=1,u_2=2$

So either $x+\frac{1}{x}=1$ or $2$

From the first solution we get

$\Rightarrow$$x^2-x+1=0$ which has no solution

From the second one we get

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

$\Rightarrow$$x=1$