Math problems

[Schwalb]

1. When the expression (2+ax)^10 is expanded, the coefficient of the term in x^3 is 414720. Find the value of a.

2. The equation x^2-2kx+1=0 has two distinct real roots. Find the set of all possible values of k.

Edited May 18, 2009 by Schwalb

[Toffu-san]

1. When the expression (2+ax)^10 is expanded, the coefficient of the term in x^3 is 414720. Find the value of a.

2. The equation x^2-2kx+1=0 has two distinct real roots. Find the set of all possible values of k.

I will try to answer you:

The first one:

You have to know the expansion of (x+y)^n (Newton's Binome, http://en.wikipedia.org/wiki/Binomial_theorem), then we have that:

We have that k=3, for this motive the coefficient of the term x^3 is:

2^(10-3)· a^3 · 10!/(3!·7!)=414720

Now you can find a .

For second one:

We know that in a quadratic equation the solve for a·x^2+b·x+c=0 is x=(-b+-sqrt{b^2-4ac})/2a, if this equation have to have two distinc real roots, must exist sqrt{b^2-4ac} in R and it have to be different to 0 (if it was zero it will have a double root). For this motive b^2-4ac>0. Then:

(-2·k)^2-4>0

4·k^2-4>0

k^2>1

|k|>1

I hope to have helped you.