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#1 2008-02-03 00:49:25
- ganesh
- Moderator
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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
- Moderator
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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
- Power Member
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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
- Novice
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Re: IIT level questions - 1
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:
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.
