SOLUTION: x^2-5x+35=0 it has no real solutions but what would be the irrational solution?

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Question 240628: x^2-5x+35=0
it has no real solutions but what would be the irrational solution?
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
This is a job for the quadratic formula. For
ax%5E2+%2B+bx+%2B+c+=+0
the solutions can be found with:
x+=+%28-b+%2B-+sqrt%28b%5E2+-+4ac%29%29%2F%282a%29

In your equation, x%5E2-5x%2B35
a = 1
b = -5
c = 35
So the solutions are:
x+=+%28-%28-5%29+%2B-+sqrt%28%28-5%29%5E2+-+4%281%29%2835%29%29%29%2F%282%281%29%29
Simplifying inside the square root:
x+=+%28-%28-5%29+%2B-+sqrt%2825+-+4%281%29%2835%29%29%29%2F%282%281%29%29
x+=+%28-%28-5%29+%2B-+sqrt%2825+-+140%29%29%2F%282%281%29%29
x+=+%28-%28-5%29+%2B-+sqrt%28-115%29%29%2F%282%281%29%29
At this point, since there is a negative number in the square root, we know that the solutions will be complex (not irrational). For complex numbers we use "i" which stand for sqrt%28-1%29:
x+=+%28-%28-5%29+%2B-+sqrt%28%28-1%29%28115%29%29%29%2F%282%281%29%29
x+=+%28-%28-5%29+%2B-+sqrt%28-1%29sqrt%28115%29%29%2F%282%281%29%29
x+=+%28-%28-5%29+%2B-+i%2Asqrt%28115%29%29%2F%282%281%29%29
Simplifying the rest we get:
x+=+%285+%2B-+i%2Asqrt%28115%29%29%2F2
Separating this into two terms and writing this in a+bi form we get:
x+=+5%2F2+%2B-+%28sqrt%28115%29%2F2%29i