Prove that (5-2√3)2 is irrational number?

We will prove it in a  same way as we prove that {tex} \sqrt{3}{/tex} is an irrational number..

Let us assume that {tex}({5-2\sqrt{3}})^2{/tex} is a rational number. 

Then          {tex}( {5-2\sqrt{3}})^2={p\over q}{/tex}    where p and q are co-prime.

{tex} =>{/tex}  {tex}25-2×5×2{\sqrt 3}+(2{\sqrt 3})^2={p\over q}{/tex}    [by using {tex}(a-b)^2=a^2-2ab+b^2{/tex}

                             {tex}25-20{\sqrt 3} +4×3={p\over q}{/tex}

                               {tex}25-20{\sqrt 3}+12={p\over q}{/tex}

                                         {tex}37-20{\sqrt 3}={p\over q}{/tex}

                                        {tex}37+{p\over q}=20{\sqrt3 }{/tex}

                                       {tex}{37\over 20} +{p\over 20q}={\sqrt 3}{/tex}

      Clearly L.H.S. is a sum of two rational number and therefore L.H.S  is rational.

So {tex}{\sqrt 3}{/tex} is a rational number.

But  we know that {tex}{\sqrt3}{/tex} is an irrational number.So our assumption is wrong.

Hence {tex}({5-2\sqrt{3}})^2{/tex}is an irrational number.

prove (5-2 root3)^2 ₌ 25-20root 3+12=p/q consider it as rational first

but solving it we get that root 3 could be represented as rational but it is irrational number thus our assumption that (5-2 root 3)²was wrong and hence it is irrational