x^2 + 2x + 2 = 0
I would use "completing the square"
x^2 + 2x = -2
x2 + 2x + 1 = -2+1
(x+1) = -1
x+1 = ±√-1, remember √-1 = i
x = -1 ± i
The answer has an imaginary number in it which we're supposed to solve for, but I'm having trouble with it.
Thanks for any help!
I would use "completing the square"
x^2 + 2x = -2
x2 + 2x + 1 = -2+1
(x+1) = -1
x+1 = ±√-1, remember √-1 = i
x = -1 ± i
x^2 + 2x +1 = -1
(x + 1)(x + 1) = -1
(x +1)^2 = -1
(x +1) = sr(-1) (sr squareroot of)
x = -1 + sr(-1)
x = (-b ± √(b^2 - 4ac)) / (2a), where a, b, and c are the coefficients of the quadratic equation in the form ax^2 + bx + c = 0.
In our case, a = 1, b = 2, and c = 2. Substituting the values into the quadratic formula, we get:
x = (-2 ± √(2^2 - 4(1)(2))) / (2(1))
= (-2 ± √(4 - 8)) / 2
= (-2 ± √(-4)) / 2
Since we have a negative value inside the square root, we can simplify it using the imaginary unit "i" which is defined as √(-1). Therefore,
√(-4) = 2i
Substituting the value back into the equation, we have:
x = (-2 ± 2i) / 2
= -1 ± i
Hence, the solutions to the quadratic equation x^2 + 2x + 2 = 0 are:
x = -1 + i
x = -1 - i
The quadratic formula states that for any quadratic equation ax^2 + bx + c = 0, the solutions can be found using the following formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our 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)
Simplifying further:
x = (-2 ± √(4 - 8)) / 2
x = (-2 ± √(-4)) / 2
Here, we encounter an imaginary number, since the square root of a negative number is represented by the imaginary unit "i." The square root of -4 is 2i, so our equation becomes:
x = (-2 ± 2i) / 2
Simplifying further:
x = -1 ± i
Thus, the solutions to the equation x^2 + 2x + 2 = 0 are x = -1 + i and x = -1 - i.