Show: t(x) = x√2 + √3 is irrational. Hint: consider t(x)² Earlier in the question we were asked to show that the square root of 6 is irrational, which I did. But I can't seem to figure the last part out. I have included an image for reference. 

Help is much appreciated!

Let $t(x)=x\sqrt{2}+\sqrt{3}$
$t(x)^2=2x^2+3+2x\sqrt{6}$
Suppose $t(x)$ is rational, and we know $x$ is rational. 
Hence $\dfrac{t(x)^2-2x^2-3}{2x}$ is rational
But we know $\sqrt{6}$ is not, hence contradiction. 
