How many roots, real or complex, does the polynomial $$ 7+5 x^{4}-3 x^{2} $$ have in all?

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To determine the number of roots, real or complex, of the polynomial $$ 7 + 5x^4 - 3x^2 $$, we can start by rewriting it in a more standard form. We can express the polynomial as:

$$
f(x) = 5x^4 - 3x^2 + 7
$$

This is a polynomial of degree 4, which means it can have up to 4 roots in total, counting both real and complex roots.

Next, we can analyze the polynomial further by substituting $$ y = x^2 $$. This gives us a new polynomial in terms of $$ y $$:

$$
f(y) = 5y^2 - 3y + 7
$$

Now, we will find the roots of this quadratic polynomial using the quadratic formula:

$$
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$

Here, $$ a = 5 $$, $$ b = -3 $$, and $$ c = 7 $$. We first calculate the discriminant:

$$
b^2 - 4ac = (-3)^2 - 4 \cdot 5 \cdot 7 = 9 - 140 = -131
$$

Since the discriminant is negative ($$-131 < 0$$), the quadratic $$ 5y^2 - 3y + 7 $$ has no real roots. This means that $$ f(y) $$ has 2 complex roots (since a quadratic polynomial has 2 roots in total).

Since $$ y = x^2 $$, each complex root $$ y $$ corresponds to two values of $$ x $$ (specifically, $$ x = \pm \sqrt{y} $$). However, since $$ y $$ is complex, the square roots will also be complex.

Thus, the polynomial $$ f(x) = 5x^4 - 3x^2 + 7 $$ has 2 complex roots for $$ y $$, and each of these corresponds to 2 complex roots for $$ x $$. Therefore, the total number of roots $$ x $$ is:

$$
2 \text{ (complex roots for } y) \times 2 = 4 \text{ (complex roots for } x)
$$

In conclusion, the polynomial $$ 7 + 5x^4 - 3x^2 $$ has a total of

$$
\boxed{4}
$$

roots, all of which are complex.
The polynomial $$ 7 + 5x^4 - 3x^2 $$ has 4 complex roots.

How many roots, real or complex, does the polynomial have in all?

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