Mathematics - Quadratic Equation Question with Solution | TestHub

Therefore, a necessary condition for $f(x)=0$ to be a reciprocal equation is that $0$ is not a root of it, i.e., $f(0)\ne 0$. Let $f(x)=0$ be a reciprocal equation of degree $n$ having roots ${\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n$, none of which are zero.

Let $\psi (x)=0$ be the equation whose roots are $\frac{1}{{\alpha }_1},\frac{1}{{\alpha }_2},\dots ,\frac{1}{{\alpha }_n}$.

Then the equations $f(x)=0$ and $\psi (x)=0$ are identical.

Let $a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x+a_0=0$, with $a_n\ne 0$, be a reciprocal equation. Then it is identical with the equation:

$$a_0{x}^n+a_1{x}^{n-1}+\cdots +a_{n-1}x+a_n=0$$

Let $a_0\ne 0$.

Therefore, $(a_n,a_{n-1},\dots ,a_0)=K(a_0,a_1,\dots ,a_n)$ for some $K\ne 0$.

This implies $a_n=K{a}_0$, $a_{n-1}=K{a}_1$, ..., $a_0=K{a}_n$.

This implies $K^2=1$, so $K=\pm 1$.

If $K=1$, then $a_n=a_0$, $a_{n-1}=a_1$, ..., $a_0=a_n$.

This equation is said to be a reciprocal equation of the first type.

If $K=-1$, then $a_n=-a_0$, $a_{n-1}=-a_1$, ..., $a_0=-a_n$.

This equation is said to be a reciprocal equation of the second type.

A reciprocal equation is said to be of the standard form if it is of the first type and of even degree.