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#1 2008-02-03 00:49:25

ganesh
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IIT level questions - 1

Indian Institute of Technology level questions....

1. Find all those roots of the equation


whose imaginary part is positive.

2. If the roots of the equation ax²+bx+c=0 be in the ratio m:n, prove that


Character is who you are when no one is looking.
 

#2 2008-02-03 01:12:57

mathsyperson
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Re: IIT level questions - 1

Let x = z6.

Then x² - 56x - 512 = 0
∴ x = 64 or -8, by the quadratic equation.
∴ z6 = 64 or -8.

Consider the case when z6 = 64.





sin(kπ/3) is positive for k = 1 or 2, so in this case, z ∈ {1+i√3, i√3-1}.

In the case where z6 = -8, we can use similar reasoning to show that:



This time, the appropriate k-values are  4 and 5, and so we get that z ∈ {[√(2)+i√(6)]/4, [-√(2)+i√(6)]/4}.

Those four values of z make up the complete solution.


Why did the vector cross the road?
It wanted to be normal.
 

#3 2008-02-03 22:15:57

Kurre
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Re: IIT level questions - 1

2:
the roots x1 and x2 can be written as:


for a constant k
According to Viètes identity:


i.e.


substituting b and c in the original equation yields:

 

#4 2008-03-15 16:02:31

ayush007
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Re: IIT level questions - 1

big_smile

mathsyperson wrote:

Let x = z6.

Then x² - 56x - 512 = 0
∴ x = 64 or -8, by the quadratic equation.
∴ z6 = 64 or -8.

Consider the case when z6 = 64.





sin(kπ/3) is positive for k = 1 or 2, so in this case, z ∈ {1+i√3, i√3-1}.

In the case where z6 = -8, we can use similar reasoning to show that:

big_smilekiss

This time, the appropriate k-values are  4 and 5, and so we get that z ∈ {[√(2)+i√(6)]/4, [-√(2)+i√(6)]/4}.

Those four values of z make up the complete solution.

 

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