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 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 $\displaystyle (1-2i)x+7-8i=0$. x-2ix+7-8i =0 ; Let x=a+bi ; a+bi-2i(a+bi)-8i+7=0 ; a+7-2b +i(2a- 8=0) . Now solve the system of equations ,{ a+7=2b ; 2a=8 } . a=4 , b=11/2 . The solution is $\displaystyle x=a+bi =4+\frac{11}{2} i$. Thanks from topsquark 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? Thanks from topsquark
August 22nd, 2019, 09:57 PM   #4
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From: România

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Quote:
 Originally Posted by skipjack Was the equation typed correctly?
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? Thanks from topsquark
August 23rd, 2019, 08:13 PM   #6
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From: România

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Quote:
 Originally Posted by skipjack 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?
Yes , will approximate complex solutions suffice.Thank you very much!

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
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Quote:
 Originally Posted by skipjack -1.7570083... + 0.20707294...$i$, -0.32214108... - 1.52875584...$i$, -0.12085059... + 1.71880263...$i$
Hello,

How do we get to these solutions without using Cardan's formulas?Thank you very much!

All the best,

Integrator

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