Show: t(x) = x√2 + √3 is irrational. Hint: consider t(x)²

Question

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!

Answer 1

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.

Answer 2

Of course you have $$t(x)^2 = 2x^2+2x\sqrt{6}+3.$$

Now $x$ is rational but $\sqrt{6}$ is irrational and so what can you conclude about the rationality/irrationality of $t(x)^2$?

But now suppose $t(x)$ is rational...