Biquadratic equations

An example of a biquadratic equation is 3x⁴ + 7x² + 5 = 0. In general, we call an equation biquadratic if it can be rearranged, by doing the same to both sides, into the form

$$a{x}^4+b{x}^2+c=0$$

where a, b and c are real numbers and the coefficient a is not zero. The equation looks a little like a quadratic equation, except that the powers are twice as big. How do we solve such an equation? By using a small trick to turn it into a quadratic equation. The trick is called substitution. We make the substitution by saying that t = x² and now, everywhere the original equation had x², we write t. As an example, take the biquadratic equation

$$x^4-2x^2-15=0$$

We replace every x² with t, remembering that x⁴ is really x² · x² and so x⁴ = t²:

$$t^2-2t-15=0$$

Now we can see why we made this substitution: we have an ordinary quadratic equation. We can solve it, for example, by working out the discriminant or by completing the square. The discriminant is

$$D=b^2-4ac=(-2{)}^2-4\cdot 1\cdot (-15)=64$$

and so the roots of the equation are

$$t_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{2\pm \sqrt{64}}{2}=\frac{2\pm 8}{2}$$

We get two roots:

$$t_1=5,t_2=-3$$

At this point we know the solutions of the quadratic equation t² − 2t − 15 = 0. What remains is to find the solutions of the original biquadratic equation. We know that if t equals 5 or minus 3, the equation is satisfied. But we don’t want to know the solutions for t; we want the solutions for x. We know that t = x², because that is the substitution we chose. So if we now solve the equations

$$x^2=5\text{and}{x}^2=-3$$

we get the solutions of the original equation for x. The first equation is easy: if x² = 5, then

$$x_1=\sqrt{5},x_2=-\sqrt{5}$$

The second equation is harder, because x² = −3 has no solution in the real numbers; we would have to find the solutions in the complex numbers. If we are looking only for real solutions, we can say that the original biquadratic equation has two real roots, $x_1=\sqrt{5}$ and $x_2=-\sqrt{5}$.