SOLUTION: I am trying to solve this problem using the quadratic formula but I need help. here is the problem: x^2+12=0

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I am trying to solve this problem using the quadratic formula but I need help. here is the problem: x^2+12=0       Log On


   

Question 996252: I am trying to solve this problem using the quadratic formula but I need help. here is the problem:
x^2+12=0
Found 3 solutions by Alan3354, fractalier, LinnW:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
I am trying to solve this problem using the quadratic formula but I need help. here is the problem:
x^2+12=0
--------
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B0x%2B12+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%280%29%5E2-4%2A1%2A12=-48.

The discriminant -48 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -48 is + or - sqrt%28+48%29+=+6.92820323027551.

The solution is , or
Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B0%2Ax%2B12+%29

=================
x^2+12=0
x^2 = -12
x = ħsqrt(-12) = ħ2sqrt(3)i

Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
For x^2+12=0, the a=1, b=0, and c=12.
Thus, from the quadratic formula,
x = [-0 +- SQRT(0^2 - 4(1)12)] / 2
x = +- SQRT(-48) / 2
x = +- 4i*SQRT(3) / 2
x = +- 2i*SQRT(3)

Answer by LinnW(1048) About Me  (Show Source):
You can put this solution on YOUR website!
First the quadratic formula
%28-b+%2B-+sqrt%28b%5E2-4ac%29%29%2F2a
Our equation is x%5E2%2B12=0
The a in the formula is the coefficient of the square term
so for our equation, a = 1
Since our equation does not have a middle term, b = 0
c is the value of the constant which for us is 12
Substituting we get
%28-0+%2B-+sqrt%280-4%281%29%2812%29%29%29%2F2%281%29
%28-0+%2B-+sqrt%28-48%29%29%2F2
%28-0+%2B-+sqrt%28-12%2A4%29%29%2F2
%28-0+%2B-+2%2Asqrt%28-12%29%29%2F2
%28-0+%2B-+sqrt%28-12%29%29
%28%2B-+sqrt%28-12%29%29
Since we have the square root of a negative number we have no real root
Notice that we could have solved more easily.
Take our equation, x%5E2%2B12=0
add -12 to each side
x%5E2=-12
Take the square root of each side and we get
x=%2B-sqrt%28-12%29