Biquadratic Equations
A biquadratic equation is a fourth-degree trinomial of the form $$ ax^4 + bx^2 + c = 0 $$
Biquadratic equations can be solved using the same approach as trinomial equations.
To simplify the problem, introduce an auxiliary variable z = x2, which reduces the equation to a quadratic form:
$$ az^2 + bz + c = 0 $$
Once the roots z1 and z2 of the quadratic equation are found, the corresponding values of x follow directly from the relation z = x2:
$$ x_1 = \pm \sqrt{z_1} $$
$$ x_2 = \pm \sqrt{z_2} $$
Worked Example
Consider the following biquadratic equation:
$$ x^4 - 7x^2 - 144 = 0 $$
Substitute z = x2 to obtain:
$$ z^2 - 7z - 144 = 0 $$
This is now a quadratic equation.
Since the discriminant is positive, the equation has two distinct real roots.
$$ \Delta = b^2 - 4ac = (-7)^2 - 4(1)(-144) = 49 + 576 = 625 $$
Compute z1 and z2 using the quadratic formula:
$$ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$
$$ z = \frac{-(-7) \pm \sqrt{625}}{2} $$
$$ z = \frac{7 \pm 25}{2} $$
$$ z = \begin{cases} \frac{7 - 25}{2} = -9 \\ \\ \frac{7 + 25}{2} = 16 \end{cases} $$
Thus, z1 = -9 and z2 = 16. From z = x2 we can determine the corresponding values of x:
$$ \begin{cases} z_1 = x_1^2 \\ \\ z_2 = x_2^2 \end{cases} $$
$$ \begin{cases} -9 = x_1^2 \\ \\ 16 = x_2^2 \end{cases} $$
Taking the square root of both sides gives:
$$ \begin{cases} \sqrt{-9} = \sqrt{x_1^2} \\ \\ \sqrt{16} = \sqrt{x_2^2} \end{cases} $$
Hence:
$$ \begin{cases} x_1 = \sqrt{-9} \\ \\ x_2 = \pm 4 \end{cases} $$
The first solution, x1 = √(-9), is not valid in the real number system since the radicand is negative and no real number squared yields a negative result.
Therefore, the real solutions of the original equation x4 - 7x2 - 144 = 0 are:
$$ x = \pm 4 $$
This completes the solution.
