Hi

Finding the solution set in a +bi form:

6x^2 + 4x +1 =0

 a=6, b=4 and c=1  |Yes!



x = -b  ± sqrt(b^2-4ac)]/12   |Yes! Finish substituting for b, b^2 and 4ac

x = -4 ± sqrt(16-24)]/12

x = -4 ± sqrt(-8)]/12

x = -4 ± sqrt(-1*4*2)]/12    |sqrt(-1) = i

x = -4 ± 2isqrt(2)]/12

x = -1/3 ± (1/6)isqrt(2)

x = -1/3 + (1/6)isqrt(2) x = -1/3 - (1/6)isqrt(2)

  

Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
 


 
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 -8 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 -8 is + or - . 

  

      The solution is , or

  Here's your graph:





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