xsquared+1=x?

Answer 1

x^2 +1 = x

Put in quadratic form..
x^2 -x +1= 0

The determinant, b^2 - 4ac is negative, (-3), so there is no real solution for x.

Plugging into the quadratic formula, we get..
(1 +- sqrt(1 - 4)) /2

Giving the complex roots
x = (1 + i*sqrt(3))/2
x = (1 - i*sqrt(3))/2

Answer 2

This converts to x^2 - x + 1 = 0
This will not factor, and has no real solutions.
Using the quadratic formula :

{1 +- sqrt(1 - 4)} / 2 or {1 +-sqrt(-3)} / 2

This is one half plus half of the square root of negative three, and one half minus half of the square root of negative three.

These are imaginary numbers and can also be stated as 1/2 + 3i/2 and 1/2 - 3i/2

Answer 3

a million-3/(x^2) = a million/x -3/(x^2) = a million/x - x/x -3 = ((a million-x)/(x)) * (x^2)) ( multiply x^2 to the two aspects and one x is canceled so no x on backside and that i made the a million an x/x in case you probably did no longer word) -3 = (a million-x)*x -3 = x-x^2 0 = -x^2 + x + 3 you could now do the quadratic formula :)))

Answer 4

x^2+1 = x
x^2 - x + 1 = 0
Use the quadratic equation now. a = 1, b = -1, c = 1

Answer 5

x^2 +1=x
x^2 - x + 1 = 0
D = b^2 - 4ac = (-1)^2-4*1 = -3
If D < 0,
this equation has no real solution.
The solution is :
(1 + 3i)/2
or
(1 - 3i)/2
From the x = [-b +- (root of (b^2-4ac))]/2a

Answer 6

x^2+1=x

Subtract x from both sides:
x^2-x+1=0

Use the quadratic formula:
x=-b±√b^2-4ac/2a
a=1,b=-1,c=1
x=(1±√(-1)^2-4(1*1))/2(1)
x=(1±√1-4)/2
x=(1±√-3)/2

There is no real solution unless you use imaginary numbers.

Answer 7

x² - x + 1 = 0
x = [1 ± √(1 - 4)] / 2
x = [1 ± √(- 3)] / 2
x = [1 ± i √(3)] / 2
x = [1/2 ± i √3/2]
x = 0.5 ± i ( 0.866 )

Answer 8

it gets closer as x approaches infinity