# Binomial Series

• May 27th 2012, 06:33 AM
britmath
Binomial Series
Hi,

I've come across a problem where I know how to do the middle and end of this problem, but I'm a little stuck on the beginning!

I'm told that the x and x2 terms in the expansion of (1 + px +qx2)-2 are 4 and 14 respectively, and I need to find p and q. Obviously I need to get the function into the form [(1 + rx)(1 + sx)]-2 (or similar) to perform the binomial expansion(s). Initially I tried to use the quadratic formula to solve for x in terms of p and q, and from here work out what r and s would be, but this quickly became very unwieldy, and I couldn't successfully simplify the expansion. I feel like I'm missing something really simple in the factorisation of (1 +px +qx2); is there a really simple step I'm missing, or is my method correct?

Thanks very much!
• May 27th 2012, 06:43 AM
Prove It
Re: Binomial Series
\displaystyle \displaystyle \begin{align*} q\,x^2 + p\,x + 1 &= q\left(x^2 + \frac{p}{q}\,x + \frac{1}{q}\right) \\ &= q\left[x^2 + \frac{p}{q}\,x + \left(\frac{p}{2q}\right)^2 - \left(\frac{p}{2q}\right)^2 + \frac{1}{q}\right] \\ &= q\left[\left(x + \frac{p}{2q}\right)^2 - \frac{p^2}{4q^2} + \frac{4q}{4q^2}\right] \\ &= q\left[\left(x + \frac{p}{2q}\right)^2 - \left(\frac{p^2 - 4q}{4q^2}\right)\right] \\ &= q\left[\left(x + \frac{p}{2q}\right)^2 - \left(\frac{\sqrt{p^2 - 4q}}{2q}\right)^2\right] \\ &= q\left(x + \frac{p - \sqrt{p^2 - 4q}}{2q}\right)\left(x + \frac{p + \sqrt{p^2 - 4q}}{2q}\right)\end{align*}

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