Question

Pls solve x^²+2x+2?

Answer 1

This equation does not have a 'real' solution.
`x^² + 2x + 2 = 0`
`x = (-2 ± 2 i )/2` where i `= sqrt[-1]`

Explanation:

First we "factor" it. This is done by making two factors (for a quadratic like this) and finding the correct coefficients.

`x^² + 2x + 2 = 0` ; `(x ? a)(x ? b)` from this form you can see that we need the constants to be:
`x^² + (xa + xb) + ab`; or `x^² + x(a + b) + ab`
So, ab = 2 and a + b = 2; a = 2 - b
This can't be solved by inspection (looking at it) so we will need to use the quadratic formula. We now have the equation in the form of a quadratic, and can solve it by using the quadratic formula. See http://www.purplemath.com/modules/quadform.htm for instructions.
For `ax^2 + bx + c = 0`, the values of x which are the solutions of the equation are given by:

x = ​(−b±√[​b​^2​​−4ac])/2a

In this case, a = 1, b = 2 and c = 2
`x = ​(−2 ± sqrt[(2^​2​ ​− 4*1*2]))/(2*1)`
`x = ​(-2 ± sqrt[(4 – 8)])/2` ; `x = (-2 ± sqrt[-4])/2`
The negative square root indicates that this expression does NOT have 'real' root.
`x = (-2 ± 2 i )/2` where i `= sqrt[-1]`