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avatar+261 

\(x^2 -2x + 3 - 2i\sqrt{3} =0\)

 

Hello! How would I go about solving this particular equation? Factoring this seemed awful, I tried factoring out sqrt(3) out of 3 and 2i*sqrt(3) but ended up with nothing useful since the i is still there. Using the quadratic formula gave me a complex number in the root sign and I'm not sure how to handle that. What about if the x^2 had a i attached to it? So this one for instance:

 

\(ix^2 -2x + 3 - 2i\sqrt{3} =0\)

 

Thanks!

Quazars  Oct 12, 2017
 #1
avatar+87571 
+1

x^2  + 2x  +  3  -2i√3  = 0          rearrange as

 

x^2  +  2x   =   -3 + 2i√3           

 

Take   1/2 the coefficient on x = 1......square it  = 1   and add to both sides

 

x^2 + 2x + 1  =  -3  + 2i√3  + 1       factor the left side....simplify the right side

 

( x + 1)^2  = -2 + 2i√3   factor the right side

 

(x + 1)^2   =  2 (i√3 - 1)     take both roots

 

x + 1  =  ±√ [ 2  (i√3 - 1) ]     subtract 1 from both sides

 

x  =    ±√ [ 2  (i√3 - 1) ]  -   1

 

 

cool cool cool

CPhill  Oct 12, 2017
 #2
avatar+261 
0

Oh... is there no way to simplify that? It seems like a very messy answer. How about the second case with an i coefficient infront of the x^2 term? thanks

Quazars  Oct 13, 2017

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