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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 15,837

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

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

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

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:

:DThis 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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