Question

The root of the reciprocal equation of the first type and of odd degree is:
A. $x = 1$
B. $x = - 1$
C. $x = \pm 1$
D. $x = 0$

Answer 1

Hint:
Reciprocal equations of first type refer to the equations having coefficients from one end are equal to the coefficients from the other end. They are equal in both sign and magnitude. Therefore, take a polynomial with odd degree, for example, $a{x^5} + b{x^4} + c{x^3} + c{x^2} + bx + a = 0$ , where the magnitude and sign of terms from one end is equal to other end. Then, each of the given values from the option to determine the correct answer.

Complete step by step solution:
Reciprocal equations of first type are the equations having coefficients from one end are equal to the coefficients from the other end. They are equal in both sign and magnitude.
Also, any polynomial with odd degree has even terms.
Let $a{x^5} + b{x^4} + c{x^3} + c{x^2} + bx + a = 0$ be a reciprocal equation of first type, where the degree of the polynomial is odd and the number of terms is even.
Also, here the magnitude and sign of terms from one end is equal to other end.
We will find the roots by trial and error method.
Substitute $x = 1$ in the above equation,
$a{\left( 1 \right)^5} + b{\left( 1 \right)^4} + c{\left( 1 \right)^3} + c{\left( 1 \right)^2} + b\left( 1 \right) + a = 2a + 2b + 2c$
Definitely $a$ cannot be 0 otherwise the degree of the polynomial will be 4 and it will contradict our assumption.
Next, let $x = - 1$
$
  a{\left( { - 1} \right)^5} + b{\left( { - 1} \right)^4} + c{\left( { - 1} \right)^3} + c{\left( { - 1} \right)^2} + b\left( { - 1} \right) + a \\
   = - a + b - c + c - b + a \\
   = 0 \\
$
This implies, $x = - 1$ is a root of the reciprocal equations of first type.
Now, substitute $x = 0$
$a{\left( 0 \right)^5} + b{\left( 0 \right)^4} + c{\left( 0 \right)^3} + c{\left( 0 \right)^2} + b\left( 0 \right) + a = a$
This implies $x = 0$ is not the root of the equation.

Hence, option B is correct.

Note:
If the degree of the polynomial is $n$, then the polynomial is written as $a{x^n} + b{x^{n - 1}} + c{x^{n - 1}} + ..... = 0$, where $a \ne 0$. Here, we have seen that the root of the reciprocal equations of first type and of odd degree is $ - 1$. Reciprocal equations of the second type are equations having coefficients from one end equal in magnitude and opposite in sign from the coefficients from the other end. The root of reciprocal equations of second type and of odd degree is $x = 1$.