SOLUTION: It is possible for a quadratic equation to have no real-number solutions. Solve t2 + 10 = 6t

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Question 387441: It is possible for a quadratic equation to have no real-number solutions.
Solve
t2 + 10 = 6t
Found 3 solutions by Alan3354, ptaylor, ewatrrr:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Yes, it's possible. eg, x^2 + 4 = 0
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t^2 + 10 = 6t
t^2 - 6t + 10 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-6x%2B10+=+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=%28-6%29%5E2-4%2A1%2A10=-4.

The discriminant -4 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 -4 is + or - sqrt%28+4%29+=+2.

The solution is x%5B12%5D+=+%28--6%2B-i%2Asqrt%28+-4+%29%29%2F2%5C1+=++%28--6%2B-i%2A2%29%2F2%5C1+, or
Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-6%2Ax%2B10+%29

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t = 3 ± i
No real number solutions.

Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!
t^2-6t+10=0
t+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
t+=+%286+%2B-+sqrt%28+36-40+%29%29%2F%282%29+
t+=+%286+%2B-+sqrt%28+-4+%29%29%2F%282%29+
t+=+%286+%2B-2i%29%2F%282%29+
t+=+%286+-2i%29%2F%282%29+=t+=+%283+-i%29
t+=+%286+%2B2i%29%2F%282%29+=t+=+%283+%2Bi%29
i think so
Hope this helps---ptaylor


Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi,   

 Yes, when b^2 - 4ac < 0 in the quadratic equaton ax^2 + bx + c = 0 

that will result in sqrt%28b%5E2+-+4ac+%29 being an imaginary number.

solving using the quadratic equation

t^2 - 6t + 10 = 0

  t+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

  t+=+%286+%2B-+sqrt%28+-4%29%29%2F%282%29+

  t+=+%286+%2B-+2i%29%2F%282%29+

  t+=+3+%2B-+i%29+

this quadraic equation has no real-number solutions