SOLUTION: 5x^4 + 14x^2 - 3 = 0

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Question 190698: 5x^4 + 14x^2 - 3 = 0
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
5x^4 + 14x^2 - 3 = 0
This is a quadratic in x^2. Use the quadratic equation to find 2 zeroes for x^2, then get the sqrt of those.
(5x^2 - 1)*(x^2 + 3) = 0
x^2 = 1/5
x = +sqrt(5)/5
x = -sqrt(5)/5
--------------
x^2 = -3
x = +3i
x = -3i
-----------
x^2 + 3 = 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%2B3+=+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%2A3=-12.

The discriminant -12 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 -12 is + or - sqrt%28+12%29+=+3.46410161513775.

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%2B3+%29