x^2-x+5=0
maybe i should have put this under Algebra or just math..I have no clue.
x = [1 +/-sqrt(-19)]/2
= 0.5 +/- 2.17945 i
Both roots are complex.
So this would be written out as:
1+√19/2 and 1-√19/2
is that correct
Get to know the "quadratic formula" for problems like this. The formula is at
http://www.mathsisfun.com/quadratic-equation-solver.html
(You can also plug in coefficients and get answers directly there).
It should be also in your textbook.
To solve the equation x^2 - x + 5 = 0, you can use the quadratic formula or complete the square method.
Method 1: Quadratic Formula
The quadratic formula states that for any quadratic equation of the form ax^2 + bx + c = 0, the solutions are given by:
x = [-b ± √(b^2 - 4ac)] / (2a)
For the given equation x^2 - x + 5 = 0:
a = 1, b = -1, and c = 5
Substituting these values into the quadratic formula:
x = [-(-1) ± √((-1)^2 - 4(1)(5))] / (2(1))
x = [1 ± √(1 - 20)] / 2
x = [1 ± √(-19)] / 2
Since the discriminant (b^2 - 4ac) is negative in this case, the solutions will be complex numbers.
Method 2: Completing the Square
To complete the square, first rearrange the equation:
x^2 - x = -5
Now, add a term to both sides to complete the square on the left side. Take half of the coefficient of 'x', square it, and add it to both sides:
x^2 - x + (1/2)^2 = -5 + (1/2)^2
(x - 1/2)^2 = -5 + 1/4
(x - 1/2)^2 = -19/4
Taking the square root of both sides, you get:
x - 1/2 = ± √(-19/4)
x = 1/2 ± (√19/4)i
x = (1 ± √19i) / 2
Therefore, the solutions to the equation x^2 - x + 5 = 0 are complex numbers: x = (1 ± √19i) / 2.