Surds help

Show positive square root of (7+4sq rt3)is (sq rt3+2)

Firstly, by saying 'positive square root' it may seem as if $\sqrt49 = \pm7$ but it's important to understand that $\sqrt{x^2} \equiv x$. So saying positive square root is the same thing as saying square root because square roots are always positive .

If you want to prove the above, try working from the left, squaring both sides and see what you get.

"This can be written as 4+3+ (2*2*√3)
Then this becomes the square of 2+√3.Here we have used (a + b)^2 = a^2 + 2ab + b^2 "Haven't learned this yet but this is what I've come up with from Googling lol

...but it's important to understand that $\sqrt{x^2} \equiv x$....

If you want to prove the above, try working from the left, squaring both sides and see what you get.

"This can be written as 4+3+ (2*2*√3)
Then this becomes the square of 2+√3.Here we have used (a + b)^2 = a^2 + 2ab + b^2 "Haven't learned this yet but this is what I've come up with from Googling lol

Firstly, by saying 'positive square root' it may seem as if $\sqrt49 = \pm7$ but it's important to understand that $\sqrt{x^2} \equiv x$. So saying positive square root is the same thing as saying square root because square roots are always positive .

Not quite the case. Take $x=-7$ then $\sqrt{x^2} = \sqrt{(-7)^2}=7 \neq x =-7$

In fact, $\sqrt{x^2} \equiv |x|$ which is a positive quantity for any $x$

Changed to $\sqrt{x^2} \equiv x, x > 0$. I assume this is mathematically correct?

Yes.

$|x|$ can be defined as:

$x$ if $x \geq 0$
$-x$ if $x <0$

You just stated the first condition pretty much.