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#1 2008-02-02 01:49:25

ganesh
Moderator
Registered: 2005-06-28
Posts: 13,853

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.

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#2 2008-02-02 02:12:57

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: IIT level questions - 1

Let x = z[sup]6[/sup].

Then x² - 56x - 512 = 0
∴ x = 64 or -8, by the quadratic equation.
∴ z[sup]6[/sup] = 64 or -8.

Consider the case when z[sup]6[/sup] = 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 z[sup]6[/sup] = -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.

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#3 2008-02-02 23:15:57

Kurre
Member
Registered: 2006-07-18
Posts: 280

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:

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#4 2008-03-14 17:02:31

ayush007
Member
Registered: 2008-03-14
Posts: 1

Re: IIT level questions - 1

big_smile

mathsyperson wrote:

Let x = z[sup]6[/sup].

Then x² - 56x - 512 = 0
∴ x = 64 or -8, by the quadratic equation.
∴ z[sup]6[/sup] = 64 or -8.

Consider the case when z[sup]6[/sup] = 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 z[sup]6[/sup] = -8, we can use similar reasoning to show that:

:Dkiss

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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