Question

50 points !!!!!!!

$4 + 2\sqrt{3}$
kese hua ye bta do​

Answer 1

Let 4 + 2 root 3 be rational.

$= > 4 + 2 \sqrt{3} = \binom{p}{q} \\ = > 2\sqrt{3} = \binom{p - 4q}{q} \\ = > \sqrt{3} = \binom{p - 4q}{2q} \\ since \: (p - 4q) \: and \: 2q \: are \: prime \: numbers \\ = > \binom{p - 4q}{2q} is \: rational \\ = > \sqrt{3} \: is \: also \: rational \\ but \: \sqrt{3 \: } is \: irrational \\ = > \binom{p - 4q}{2q} \: is \: also \: irrational$

Answer 2

$\huge\boxed{Question}$

$4 + 2 \sqrt{3}$

$\huge{\pink {\sf\underline{Answer}}}$

=] 7-√3-x = 3+√3

now taking all the constant one side and variable one side we get

=] -x = 3+ √3-7+√3

=] -x= -7 +3+√3+√3

So,

-x = -4+2√3

negative sign of X will transfer to RHS we get

=] x = -(-4+2√3)

=] x = 4-2√3

thanks ♥️