How do you find the zeroes of f (x) =5x^4 − 2x^2 − 3?

1 Answer
May 3, 2015

This function is an example of a bi-quadratic function, which is the polynomial function of the 4^(th) degree with no terms of an odd degree.

The general polynomial of the 4^(th) degree looks like this:
f(x)=a_0x^4+a_1x^3+a_2x^2+a_3x^1+a_4x^0
Since no odd degree terms are present, general expression for a bi-quadratic function is:
f(x)=a_0x^4+a_2x^2+a_4x^0

Finding the values of an unknown x where this function equals to zero is a simple three-step procedure.

Step 1. Substitute y=x^2. Then the equation f(x)=0 that determines the zeros of a function is transformed into an equation with an unknown y:
a_0y^2+a_2y+a_4=0

Step 2. The above equation is a regular quadratic equation that we know how to solve. Its two solutions are:
y_1=(-a_2+sqrt(a_2^2-4a_0a_4))/(2a_0)
y_2=(-a_2-sqrt(a_2^2-4a_0a_4))/(2a_0)
(solutions might not be real if a_2^2-4a_0a_4<0, they are supposed to be discarded).

Step 3. Knowing the value of an unknown y (actually, from zero up to two values, depending on coefficients), we can find up to four values of x since y=x^2:
x_1=sqrt(y_1); x_2=-sqrt(y_1); x_3=sqrt(y_2); x_4=-sqrt(y_2);
(depending on the coefficients, certain solutions might not be real)

I think it would be useful for a student who ask this question to do the math with concrete coefficients given in the problem.

As an illustration, here is a graph of the given function that shows where it takes zero values. It shows that this function has only two real values of x where it equals to zero, x=1 and x=-1, which implies that one of the solutions of an equation for y is negative and there is no x that would be equal to it if raised to a power of 2.
graph{5x^4-2x^2-3 [-3, 3, -4, 4]}