SOLUTION: I'm suppose to solve: x=square root of (4x-21)

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Question 448755: I'm suppose to solve:
x=square root of (4x-21)
Answer by jorel1380(3719) About Me  (Show Source):
You can put this solution on YOUR website!
x=√4x-21
x2=4x-21
x2-4x+21=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-4x%2B21+=+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=%28-4%29%5E2-4%2A1%2A21=-68.

The discriminant -68 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 -68 is + or - sqrt%28+68%29+=+8.24621125123532.

The solution is

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-4%2Ax%2B21+%29