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)   (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 (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

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 - .

The solution is , or
Here's your graph:

---------------
t = 3 ± i
No real number solutions.
Answer by ptaylor(2198)   (Show Source): You can put this solution on YOUR website!
t^2-6t+10=0




=
=
i think so
Hope this helps---ptaylor


Answer by ewatrrr(24785)   (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  being an imaginary number.

solving using the quadratic equation

t^2 - 6t + 10 = 0

  

  

  

  

this quadraic equation has no real-number solutions 

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