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**RauLiTo****Member**- From: Bahrain
- Registered: 2006-01-11
- Posts: 142

get X and Y in the below equation :

X (X+iY)² - 10(X+iY) + 14 = 0

help please guys i am getting crazy

ImPo$$!BLe = NoTH!nG

Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...

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**RauLiTo****Member**- From: Bahrain
- Registered: 2006-01-11
- Posts: 142

guys ?

ImPo$$!BLe = NoTH!nG

Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

if that first x wasnt there, you could use quadratic equation, but since it is, i dont know

The Beginning Of All Things To End.

The End Of All Things To Come.

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**RauLiTo****Member**- From: Bahrain
- Registered: 2006-01-11
- Posts: 142

:d anybody else ?

ImPo$$!BLe = NoTH!nG

Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...

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**RauLiTo****Member**- From: Bahrain
- Registered: 2006-01-11
- Posts: 142

come on guys !!! is my question is not clear or what ?

ImPo$$!BLe = NoTH!nG

Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...

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**tom112****Guest**

First, expand the brackets...

x(x + iy)^2 becomes x(x^2 + 2xyi - y^2)

which in turn becomes x^3 + 2(x^2)yi - xy^2

the second term expands to -10x - 10iy

So the equation with the brackets expanded reads

x^3 + 2(x^2)yi - xy^2 -10x - 10iy + 14 = 0

Simplify like terms and generally tidy things up:

x^3 -xy^2 - 10x + 14 + i(2(x^2)y - 10y) = 0 (+ 0i), this makes the next step slightly clearer...

Equate real and imag. coeffs:

x^3 -xy^2 - 10x + 14 = 0

=> y = (x^2 - 1 - 10 + 14)^1/2

=> y = (x^2 + 3)^1/2 (Not very pretty...)

2(x^2)y - 10y = 0

=> x^2 - 5 = 0

=> x = + or - 5^1/2

The + or - bit becomes irrelevant, since the x is squared in the previous equation.

=> y = (5 + 3)^1/2 = sqrt(8) (supposed to be a 9?)

So your solutions (possibly) are x = +/- sqrt(5), y = sqrt(8)

I highly doubt I have gone straight through this without making some sort of error, so please check it - the idea was to give you and idea of which process to undertake when solving this questions.

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