August 20th, 2019, 07:01 AM  #1 
Member Joined: Aug 2018 From: România Posts: 88 Thanks: 6  An equation
Hello all, Solve the equation $\displaystyle (1+2i)x^3+(3+4i)x^2+(5+6i)x+7+8i=0$ where $\displaystyle i ^ 2 = 1$. All the best, Integrator 
August 20th, 2019, 08:17 AM  #2 
Senior Member Joined: Dec 2015 From: somewhere Posts: 647 Thanks: 93 
Substitute for $\displaystyle x=a+bi$ and then expand the polynomial. Now find a and b . For example :equation $\displaystyle (12i)x+78i=0$. x2ix+78i =0 ; Let x=a+bi ; a+bi2i(a+bi)8i+7=0 ; a+72b +i(2a 8=0) . Now solve the system of equations ,{ a+7=2b ; 2a=8 } . a=4 , b=11/2 . The solution is $\displaystyle x=a+bi =4+\frac{11}{2} i$. Last edited by idontknow; August 20th, 2019 at 08:31 AM. 
August 20th, 2019, 10:19 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,975 Thanks: 2228 
Was the equation typed correctly?

August 22nd, 2019, 09:57 PM  #4 
Member Joined: Aug 2018 From: România Posts: 88 Thanks: 6  Hello, I do not understand! Which equation is not correct? Thank you very much! All the best, Integrator Last edited by skipjack; August 22nd, 2019 at 10:33 PM. 
August 22nd, 2019, 10:36 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,975 Thanks: 2228 
The equation in the first post has three distinct complex roots, which can be obtained using the cubic formula. They aren't real numbers. Will approximate complex solutions suffice?

August 23rd, 2019, 08:13 PM  #6  
Member Joined: Aug 2018 From: România Posts: 88 Thanks: 6  Quote:
All the best, Integrator  
August 23rd, 2019, 10:42 PM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,975 Thanks: 2228 
1.7570083... + 0.20707294...$i$, 0.32214108...  1.52875584...$i$, 0.12085059... + 1.71880263...$i$

August 24th, 2019, 09:12 PM  #8 
Member Joined: Aug 2018 From: România Posts: 88 Thanks: 6  

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