How do you find the zeroes of f(x)=5x42x23?

1 Answer
May 3, 2015

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

The general polynomial of the 4th degree looks like this:
f(x)=a0x4+a1x3+a2x2+a3x1+a4x0
Since no odd degree terms are present, general expression for a bi-quadratic function is:
f(x)=a0x4+a2x2+a4x0

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

Step 1. Substitute y=x2. Then the equation f(x)=0 that determines the zeros of a function is transformed into an equation with an unknown y:
a0y2+a2y+a4=0

Step 2. The above equation is a regular quadratic equation that we know how to solve. Its two solutions are:
y1=a2+a224a0a42a0
y2=a2a224a0a42a0
(solutions might not be real if a224a0a4<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=x2:
x1=y1;x2=y1;x3=y2;x4=y2;
(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]}