SOLUTION: x^2+4x+7=0 find all real solutions and express in a+bi

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Question 52395: x^2+4x+7=0 find all real solutions and express in a+bi
Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
x^2 + 4x + 7 = 0
x^2 + 4x = -7
(x + 2)^2 = 4 - 7 = -3
x + 2 = +- sqrt(3)i
x = -2 +- sqrt(3)i
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B4x%2B7+=+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=%284%29%5E2-4%2A1%2A7=-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

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B4%2Ax%2B7+%29