prove that $2\sqrt5 +\sqrt{11}$ is irrational how would you prove that $2\sqrt5 +\sqrt{11}$ is irrational?
I started with a proof by contradiction that assumes that $2\sqrt5 +\sqrt{11}$ is rational and therefore there exist integers $a$ and $b$ such that $\frac{a}{b}=2\sqrt5 +\sqrt{11}$ and squaring both sides yields
$\frac{a^2}{b^2}=31 +4\sqrt5\sqrt{11}$ and from this point im stuck as i dont know how to continue to arrive at a contradiction.

Here's a more advanced approach, with some common details to the above. 
A monic polynomial with integer coefficients:
$$ (x^2 - (2\sqrt 5 + \sqrt{11})^2)(x^2 - (2\sqrt 5 - \sqrt{11})^2) = x^4 - 62x^2 +81 $$
Now $x = 2\sqrt 5 + \sqrt{11}$ is a root of this polynomial, but by the Rational Roots Thm., any rational root would be an integer divisor of $81$.  
Thus one only needs to verify that $2\sqrt 5 + \sqrt{11}$ is not an integer.  A simple hand computation (or calculator computation) shows this positive number lies strictly between $7$ and $8$.
