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The roots of the quadratic equation $3x^2+5x+k = 0$ are $\frac{-5\pm i\sqrt{131}}{6}$.  What is $k$?

 Aug 27, 2023
 #1
avatar+129847 
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\($\frac{-5\pm i\sqrt{131}}{6}$ \)

 

We can  solve for k  in  the  discriminant  thusly

 

5^2  - 4 (3)(k)  =  131i^2             { i^2  =  -1    and sqrt (-1)  = i }

 

5^2 - 12k = -131

 

25 - 12k  = -131

 

-12k  = -131 - 25

 

-12k  =  -156

 

k = -156 / -12  =  13

 

 

cool cool cool

 Aug 27, 2023

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