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.
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
1 answer