Can someone explain me how I can solve x^2 - 2x - 2 = 0?

Question

I can do it for simple equations, but not these ones with imaginary roots. I know there is some long crappy formula for it, but can someone explain it?

Answer 1

You mean Bhaskara's formula.

For the equation ax^2 + bx + c = 0, the roots are

x = ( (-b +/- sqrt(b^2 - 4ac) ) / (2a)

If the roots are imaginary, write them as they are. For example, solving x^2 + 2x + 3 = 0 gives

x = (-2 +/- sqrt(4 - 12))/2 =
-1 +/- sqrt(-2) =
-1 + i*sqrt(2) and -1 - i*sqrt(2)

BTW, the roots of x^2 - 2x - 2 = 0 are
( 2 +/- sqrt(4 + 8) ) / 2 = 1 +/- sqrt(3).

Answer 2

*You'll need to use the Quadtratic Equation which is,

x = [- b +/- V`(b^2 - 4ac)] / 2a

First: you have three values for three variables...
a = 1
b = -2
c = -2

Sec: replace the values with the corresponding variables...

x = [- (-2) +/- V`((-2)^2 - 4(1)(-2)] / 2(1)

x = [2 +/- V`((-2)^2 - 4(1)(-2)] / 2

x = [2 +/- V`((-2)(-2) - 4(1)(-2)] / 2

x = [2 +/- V`(4 - 4(1)(-2)] / 2

x = [2 +/- V`(4 - 4(-2))] / 2

x = [2 +/- V`(4 - (-8))] / 2

x = [2 +/- V`(4 + 8)] / 2

x = [2 +/- V`(12)] / 2

x = [2 +/- V`(2*2*3)] / 2

x = [2 +/- 2 V`3] / 2

Third: you have two solutions - one is positite, the other is negative...

a. x = [2 + 2 V`3] / 2
x = 2/2 + (2 V`3)/2
x = 1 + V`3

b. x = [2 - 2 V`3] / 2
x = 2/2 - (2 V`3)/2
x = 1 - V`3

Solutions: 1 + V`3 and x = 1 - V`3

Answer 3

-b +/- the square root of b^2 - 4 (a)(c) over 2(a)

That means the opposite of b times the sqaure root of all that over 2 times a. In your case:

2 times the square root of -2^2 -4 (1)(-2) all over 2(1)

then 2 times the square root of 12 all over 2

then thats it, unless you want to take it further.

so (2sqrt.12)/2, then you can still simplify it further...up to you, depends on what your teacher or book says too.

Answer 4

x² - 2x - 2 = 0
d = -2² - 4.1.-2
d = 4 + 8
d = 12

x = (2 +/- \/12) : 2.1
x' = (2 + 2\/3) : 2
x' = 1 + \/3
x" = 1 - \/3
Solution: {x elements of R| x' = 1+\/3 and x" = 1 - \/3}
><

Answer 5

x=[-(-2)+/-sqrt(4-4(-2))]/2
=[2+/-sqrt12]/2
x=[2+/-2sqrt3]/2
x=1+sqrt3 x=1-sqrt3
these r not imaginary roots

Answer 6

Quadratic Formula ALWAYS works.


x= (-b+(or -) sqrt (b^2-4ac))/ 2a

Answer 7

ax²+bx+c=0

[solve for x using magic]

x = [ -b ± √(b² - 4ac)] / 2a

Answer 8

srry buddy im in the 6th grade and i realy realy dont noe how to do that and plus i hate math!?!?!?!??!?!?!??!