Find the roots of the function f(x)= x^2+2x+2
Determine f(x)
a) (x+1-i)(x+1+i)
b)(x+1-i sqrt of 2)(x+1+i sqrt of 2)
c)x-1+i)(x-1-i)
d)x-1+i sqrt of 2) (x-1-i sqrt of 2)
Please Help!
For all of these, use the rule
[f(x) + ai]*[(f(x) - ai} = [f(x)]^2 + a^2
where a is a real constant.
For example, the answer to (a) is
(x+1)^2 + 1 = x^2 + 2x + 2
To find the roots of the function f(x) = x^2 + 2x + 2, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = 2, and c = 2. Substituting these values into the quadratic formula, we get:
x = (-2 ± √(2^2 - 4(1)(2))) / (2(1))
This simplifies to:
x = (-2 ± √(4 - 8)) / 2
x = (-2 ± √(-4)) / 2
Since the term under the square root is negative (4 - 8 = -4), there are no real solutions.
Instead, we have complex solutions. The complex solutions come in conjugate pairs.
The complex conjugate pairs are in the form of (x + a + bi)(x + a - bi), where a and b are real constants.
Given the options:
a) (x + 1 - i)(x + 1 + i)
b) (x + 1 - i√2)(x + 1 + i√2)
c) (x - 1 + i)(x - 1 - i)
d) (x - 1 + i√2)(x - 1 - i√2)
We can see that option a) matches the form (x + a + bi)(x + a - bi), where a = 1 and b = 1.
So, the roots of the function f(x) = x^2 + 2x + 2 are (x + 1 - i) and (x + 1 + i), as stated in option a).