Question

If $\alpha$ is a root of reciprocal equation of $f\left(x\right)=0$ , then the following is also a root of $f\left(x\right)=0$

• A. $-\alpha$
• B. ${\alpha }^2$
• C. $\frac{1}{\alpha }$
• D. $2\alpha$

Answer 1

The correct option is C $\frac{1}{\alpha }$

Let us consider the following cases:

If $a=\frac{5}{2}\Rightarrow x+\frac{1}{x}=\frac{5}{2}$

$\Rightarrow x^2+1=\frac{5x}{2}$

$\Rightarrow x^2-\frac{5x}{2}+1=0$

$\Rightarrow 2x^2-5x+2=0$

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

$\Rightarrow (x-2)(2x-1)=0$

$\Rightarrow x=\frac{1}{2}$ and $x=2$

Now consider $a=2\Rightarrow x+\frac{1}{x}=2$

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

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

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

$\Rightarrow x=1$ and $x=1$

The roots of the equation are $1,1,\frac{1}{2}$ and $2$.

This is property of a reciprocal equation that is it has $a$ as a root, then it will also have $\frac{1}{a}$ as a root.