Trying to solve for U in the below set of equations.... For context this is in the development of the Binomial tree pricing model that Cox, Ross and Rubenstein developed.
$qu + (1-q)d = e^{rt} \\ qu^{2} + (1-q)d^{2} = e^{(2r+v^{2})t}$
I was advised to find a quadratic in U but my under caffeinated brain just keeps ending up with some stupid looking equations. Appreciate this is super basic.
You have the following two equations
$ q u + (1 - q) d = A $
$ q u^2 + (1 - q) d^2 = B $
where $A = e^{rt} , B = e^{(2 r + v^2)t} $
The first is an equation of a straight line, and the second is an equation of an ellipse.
Let $u' = \sqrt{q} u $, and $ d' = \sqrt{1 - q} d $ then the first equation becomes
$\sqrt{q} u' + \sqrt{1 - q} d' = A $
and the second equation becomes
$ u'^2 + d'^2 = B $
Solve for $d'$ from the first transformed equation,
$ d' = \dfrac{ A - \sqrt{q} u' } { \sqrt{1 - q} } $
Then, plug this into the second equation
$ u'^2 + \dfrac{( A - \sqrt{q} u' )^2 }{ 1 - q } = B $
which re-arranges into
$ u'^2 - 2 u' A \sqrt{q} + A^2 - (1 - q) B = 0$
which is a quadratic equation in $u'$ and can be solved using the quadratic formula. Once $u'$ is found, then $d'$ can be computed from the equation relating $d'$ and $u'$ above. And finally, $u$ and $v$ can be computed from the defining equations of $u'$ and $d'$.