Trinomial Equations
A trinomial equation of degree higher than the second, written as $$ ax^{2n} + bx^n + c = 0 $$, has powers of the unknown x such that one is exactly double the other (2n and n). This type of equation can be simplified by introducing an auxiliary variable \( z = x^n \), which transforms it into a standard quadratic form: $$ az^2 + bz + c = 0 $$
This substitution lowers the degree of the equation, allowing us to determine the roots of the auxiliary variable z using the quadratic formula:
$$ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$
Once the real roots z1 and z2 of the auxiliary variable are obtained (if they exist), we can derive the corresponding roots of x:
$$ x^n = z_1 $$ $$ x^n = z_2 $$
Note. A trinomial equation is called a biquadratic equation when n = 2, giving the form $$ ax^4 + bx^2 + c = 0 $$
A Practical Example
Consider the following eighth-degree equation:
$$ x^8 - 17x^4 + 16 = 0 $$
This is a trinomial equation because the exponents of x are one (8) and its half (4).
Let’s introduce the auxiliary variable z = x4:
$$ z^2 - 17z + 16 = 0 $$
This yields a quadratic equation in z.
The discriminant Δ is positive, so the equation has two distinct real roots:
$$ \Delta = b^2 - 4ac = (-17)^2 - 4(1)(16) = 289 - 64 = 225 $$
We can now solve \( z^2 - 17z + 16 = 0 \) using the quadratic formula.
Given the coefficients a = 1, b = -17, and c = 16:
$$ z = \frac{-b \pm \sqrt{\Delta}}{2a} $$
$$ z = \frac{17 \pm \sqrt{225}}{2} $$
$$ z = \frac{17 \pm 15}{2} $$
$$ z = \begin{cases} \frac{17 - 15}{2} = 1 \\ \\ \frac{17 + 15}{2} = 16 \end{cases} $$
The roots of the auxiliary variable are therefore z1 = 1 and z2 = 16.
We can now find the roots of x, knowing that z = x4:
$$ x = \begin{cases} x_1^4 = z_1 \\ \\ x_2^4 = z_2 \end{cases} $$
So we have:
$$ x = \begin{cases} x_1^4 = 1 \\ \\ x_2^4 = 16 \end{cases} $$
To isolate x, we take the fourth root of both sides of each equation:
$$ x = \begin{cases} \sqrt[4]{1} \\ \\ \sqrt[4]{16} \end{cases} $$
Thus, the problem reduces to evaluating two radicals.
The fourth root of 1 is ±1, since both 14 = 1 and (-1)4 = 1.
$$ x = \begin{cases} x_1 = \pm 1 \\ \\ x_2 = \sqrt[4]{16} \end{cases} $$
The fourth root of 16 is ±2, because 24 = 16 and (-2)4 = 16.
$$ x = \begin{cases} x_1 = \pm 1 \\ \\ x_2 = \pm 2 \end{cases} $$
Therefore, the original eighth-degree equation \( x^8 - 17x^4 + 16 = 0 \) has four real roots: x1 = ±1 and x2 = ±2.
We have thus solved an eighth-degree equation by applying the same methods used to solve quadratic equations.
