Determine the number and type of solutions for the following polynomial function. Make sure to include an explanation in your work to justify how you determined your answer. P(x)=−2x^4+2x^2+4
To determine the number and type of solutions for the polynomial function P(x) = -2x^4 + 2x^2 + 4, we can use the discriminant of a quadratic equation.
First, let's rewrite the given polynomial function as a quadratic equation by setting P(x) equal to zero:
-2x^4 + 2x^2 + 4 = 0
Next, let's substitute y for x^2 to simplify the equation:
-2y^2 + 2y + 4 = 0
This equation is now in quadratic form, where a = -2, b = 2, and c = 4.
Now, we can calculate the discriminant using the formula:
D = b^2 - 4ac
Plugging in the values, we get:
D = (2)^2 - 4(-2)(4)
= 4 + 32
= 36
Since the discriminant D is greater than zero (D > 0), this means that there are two distinct real solutions for the quadratic equation.
However, we need to remember that we substituted y for x^2 earlier. So, to find the number and type of solutions for the original polynomial function P(x), we need to consider the solutions for y as well as x.
Since y = x^2, we can equate the two solutions subtracted from each other:
x^2 - x^2 = 0
0 = 0
This means that the solutions for x^2 (or y) are the same. Therefore, there are also two distinct real solutions for the polynomial function P(x).
In conclusion, the polynomial function P(x) = -2x^4 + 2x^2 + 4 has two distinct real solutions.