Trinomial equations are a specific type of polynomial equations that feature three terms expressed as a combination of variables and constants in the standard form:

\[ ax^{2n} + bx^{n} + c = 0 \]

  • \( a \), \( b \), and \( c \) are numerical coefficients.
  • \( x \) is the unknown variable.

It is worth noting that if \( n = 1 \), the equation becomes a quadratic equation, which is a particular trinomial equation in the form:

\[ ax^2 + bx + c = 0 \]

A general set of steps can be followed to solve a trinomial equation in the standard form \(ax^{2n} + bx^n + c = 0\). The first step involves introducing a substitution that will transform the equation into a quadratic equation of \(y\), which is easier to solve. For instance, we can let:

\[x^n = y\]

and obtain:

\[ ay^2 + by + c = 0 \]

We can use techniques such as factoring, completing the square, or using the quadratic formula to solve the quadratic equation in \(y\). Once the solutions for \(y\) are found, they can be substituted back into the original equation with \(y = x^n\) to find the corresponding values of \(x\). It’s essential to verify that the solutions are correct by substituting them back into the original equation to ensure accuracy.

The process may seem complex, but it’s actually practical and straightforward to implement.

Eample

Solve the equation \(3x^4-7x^2 + 2 = 0\).

Let’s substitute \(y = x^2\) to transform the equation into a quadratic equation:

\[3x^2-7x + 2 = 0\]


We can use the quadratic formula to find the value of \(y\). We obtain:

\begin{align*} y &= \frac{{-(-7) \pm \sqrt{{(-7)^2-4 \cdot 3 \cdot 2}}}}{{2 \cdot 3}} \\[0.6em] &= \frac{{7 \pm \sqrt{{49-24}}}}{6} \\[0.6em] &= \frac{{7 \pm \sqrt{25}}}{6} \\[0.6em] &= \frac{{7 \pm 5}}{6} \end{align*}

So, we found:

\[y_1 = \frac{12}{6} = 2\] \[y_2 = \frac{2}{6} = \frac{1}{3}\]


Once the solution for \(y\) are found, they can be substituted back into the original equation with \(x^2 = y\) to find the corresponding values of \(x\).

For \(y = 2\) we have: \[x^2 = 2 \quad \implies \quad x = \pm \sqrt{2} \]

For \(y = \frac{1}{3}\) we have \[x^2 = \frac{1}{3} \quad \implies \quad \pm \sqrt{\frac{1}{3}} \]


It’s essential to verify that the solutions are correct by substituting them back into the original equation to ensure accuracy. To check if the given values are solutions to the equation (3x^4-7x^2 + 2 = 0), we must plug them into the equation and verify if they satisfy it.

For \( x = \sqrt{2} \):

\[ 3(\sqrt{2})^4-7(\sqrt{2})^2 + 2 = 3 \cdot 2^2-7 \cdot 2 + 2 \] \[ = 12 – 14 + 2 = 0 \]


For \( x = -\sqrt{2} \): \[ 3(-\sqrt{2})^4-7(-\sqrt{2})^2 + 2 = 3 \cdot 2^2-7 \cdot 2 + 2 \] \[ = 12 – 14 + 2 = 0 ]


For \( x = \sqrt{\frac{1}{3}} \): \[ 3\left(\sqrt{\frac{1}{3}}\right)^4-7\left(\sqrt{\frac{1}{3}}\right)^2 + 2 = 3 \cdot \left(\frac{1}{3}\right)^2 – 7 \cdot \frac{1}{3} + 2 \] \[ = \frac{3}{9}-\frac{7}{3} + 2 = 0 \]


For \( x = -\sqrt{\frac{1}{3}} \): \[ 3\left(-\sqrt{\frac{1}{3}}\right)^4-7\left(-\sqrt{\frac{1}{3}}\right)^2 + 2 = 3 \cdot \left(\frac{1}{3}\right)^2-7 \cdot \frac{1}{3} + 2 \] \[ = \frac{3}{9}- \frac{7}{3} + 2 = 0 \]

All solutions satisfy the original equation.

The solution to the equation is: \[x = \pm \sqrt{2}, \quad \pm \sqrt{\frac{1}{3}}\]