Reciprocal Equations
What is a reciprocal equation?
A reciprocal equation is a polynomial in which the coefficients of the outer terms and those symmetrically positioned with respect to the center are either identical or opposite in sign. $$ ax^3 + bx^2 + bx + a = 0 $$ $$ ax^3 + bx^2 - bx - a = 0 $$ In the first case, the symmetrically positioned terms are equal, while in the second they are opposites.
Fundamental properties of reciprocal equations
Reciprocal equations of any degree possess two fundamental properties:
- They always have either +1 or -1 as one of their roots.
- If m ≠ |1| is a root of the equation, then its reciprocal 1/m is also a root.
Why this is useful
Knowing that a reciprocal equation necessarily has 1 or -1 as a root allows us to reduce its degree and simplify it using Ruffini's rule, a standard method for polynomial division.
Note. If a reciprocal equation of even degree has opposite coefficients, the middle term is always missing because it cancels out with its opposite. For example, a fourth-degree reciprocal equation with opposite coefficients takes the form: $$ ax^4 + bx^3 - ax - b = 0 $$ Here, the middle term cx2 is absent because, by definition, each term equidistant from the ends must have a counterpart with an opposite coefficient. In this case, that counterpart is again cx2, which becomes -cx2, and the two terms cancel each other out: cx2 - cx2 = 0.
A worked example
Consider the equation:
$$ 4x^3 - 13x^2 - 13x + 4 = 0 $$
The outer and symmetrically placed coefficients are equal, so this is a reciprocal equation.
- Testing x = 1 $$ 4(1)^3 - 13(1)^2 - 13(1) + 4 = 0 $$ $$ 4 - 13 - 13 + 4 = 0 $$ $$ 8 - 26 = 0 $$ $$ -18 = 0 $$ Therefore, x = 1 is not a root.
- Testing x = -1 $$ 4(-1)^3 - 13(-1)^2 - 13(-1) + 4 = 0 $$ $$ -4 - 13 + 13 + 4 = 0 $$ $$ 0 = 0 $$ Hence, x = -1 is a root of the equation.
We now apply Ruffini's method using the root a = -1 to factorize the cubic polynomial.
$$ P(x) = (x - a) \cdot Q(x) = 0 $$
$$ P(x) = (x - (-1)) \cdot Q(x) = 0 $$
$$ P(x) = (x + 1) \cdot Q(x) = 0 $$
We now compute the quotient polynomial Q(x) for a = -1:
$$ Q(x) = \begin{array}{c|lcc|r} & 4 & -13 & -13 & 4 \\ -1 & & -4 & 17 & -4 \\ \hline & 4 & -17 & 4 & 0 \end{array} $$
Thus, Q(x) = 4x2 - 17x + 4.
$$ (x + 1) \cdot Q(x) = 0 $$
$$ (x + 1)(4x^2 - 17x + 4) = 0 $$
We already know one root, x = -1.
To determine the remaining roots, we solve the quadratic equation 4x2 - 17x + 4 = 0:
$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$
$$ x = \frac{-(-17) \pm \sqrt{(-17)^2 - 4(4)(4)}}{2(4)} $$
$$ x = \frac{17 \pm \sqrt{289 - 64}}{8} $$
$$ x = \frac{17 \pm \sqrt{225}}{8} $$
$$ x = \frac{17 \pm 15}{8} $$
$$ x = \frac{17 \pm 15}{8} = \begin{cases} x_1 = \frac{17 - 15}{8} = \frac{2}{8} = \frac{1}{4} \\ \\ x_2 = \frac{17 + 15}{8} = \frac{32}{8} = 4 \end{cases} $$
Therefore, the remaining roots are x1 = 1/4 and x2 = 4.
Observe that every root different from ±1 appears in reciprocal pairs.
In conclusion, the set of roots of the equation is:
$$ S = \{ -1 , \tfrac{1}{4} , 4 \} $$
Proof
A] Reciprocal equation with equal coefficients
Consider the general cubic equation with equal coefficients:
$$ ax^3 + bx^2 + bx + a = 0 $$
We test whether the root is +1 or -1:
- For x = 1 $$ a(1)^3 + b(1)^2 + b(1) + a = 0 $$ $$ a + b + b + a = 0 $$ $$ 2a + 2b = 0 $$ Not a root.
- For x = -1 $$ a(-1)^3 + b(-1)^2 + b(-1) + a = 0 $$ $$ -a + b - b + a = 0 $$ $$ 0 = 0 $$ Therefore, x = -1 is a root.
B] Reciprocal equation with opposite coefficients
Now consider the general cubic equation with opposite coefficients:
$$ ax^3 + bx^2 - bx - a = 0 $$
We check whether the root is +1 or -1:
- For x = 1 $$ a(1)^3 + b(1)^2 - b(1) - a = 0 $$ $$ a + b - b - a = 0 $$ $$ 0 = 0 $$
In this case, x = 1 is the root of the reciprocal equation.
Hence, for any reciprocal equation, one of its roots will always be either x = 1 or x = -1.
