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August 20th, 2019, 07:01 AM   #1
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An equation

Hello all,

Solve the equation $\displaystyle (1+2i)x^3+(3+4i)x^2+(5+6i)x+7+8i=0$ where $\displaystyle i ^ 2 = -1$.

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August 20th, 2019, 08:17 AM   #2
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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 $\displaystyle x=a+bi =4+\frac{11}{2} i$.
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Last edited by idontknow; August 20th, 2019 at 08:31 AM.
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August 20th, 2019, 10:19 AM   #3
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Was the equation typed correctly?
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August 22nd, 2019, 09:57 PM   #4
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Quote:
Originally Posted by skipjack View Post
Was the equation typed correctly?
Hello,

I do not understand! Which equation is not correct? Thank you very much!

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Last edited by skipjack; August 22nd, 2019 at 10:33 PM.
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August 22nd, 2019, 10:36 PM   #5
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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?
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August 23rd, 2019, 08:13 PM   #6
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Quote:
Originally Posted by skipjack View Post
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!

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August 23rd, 2019, 10:42 PM   #7
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-1.7570083... + 0.20707294...$i$, -0.32214108... - 1.52875584...$i$, -0.12085059... + 1.71880263...$i$
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August 24th, 2019, 09:12 PM   #8
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Quote:
Originally Posted by skipjack View Post
-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!

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