Question

find all of the zeros of the following function:
f(x)=x^4+x^2-12

Answer 1

To find all the zeros of the function \( f(x) = x^4 + x^2 - 12 \), we need to solve the equation \( f(x) = 0 \).

\( x^4 + x^2 - 12 = 0 \)

This is a quadratic equation in terms of \( x^2 \). Let's do a substitution to make it more clear. Let \( u = x^2 \), then the equation becomes:

\( u^2 + u - 12 = 0 \)

Now, we need to factor this quadratic equation:

\( u^2 + 4u - 3u - 12 = 0 \)
\( (u^2 + 4u) - (3u + 12) = 0 \)
\( u(u + 4) - 3(u + 4) = 0 \)
\( (u - 3)(u + 4) = 0 \)

So, \( u - 3 = 0 \) or \( u + 4 = 0 \), giving us:

\( u = 3 \) or \( u = -4 \)

Now, let's substitute back \( x^2 \) for \( u \):

1. \( x^2 = 3 \)
To find x, we take the square root of both sides:
\( x = \sqrt{3} \) and \( x = -\sqrt{3} \)

2. \( x^2 = -4 \)
The square root of a negative number is imaginary, so we get complex roots:
\( x = \sqrt{-4} \) and \( x = -\sqrt{-4} \)
\( x = 2i \) and \( x = -2i \)

So the zeros of the function \( f(x) = x^4 + x^2 - 12 \) are:
\( x = \sqrt{3} \), \( x = -\sqrt{3} \), \( x = 2i \), and \( x = -2i \).