(root 2 +root 5 )^{2 }

(a+b)^{2} = a^{2} + b^{2} + 2ab

:. , (root 2)^{2} + (root 5)^{2} + 2 (root 2) ( root 5 )

= 2 + 5 + 2 ( root 10 )

7 + 2 ( root 10 )

Let assume 7+ 2(root 10) is a rational no.

=> 7 + 2 (root 10) = a / b , where a and b are co - prime no.s , where b ( not equal to) 0

=> 2 (root 10) = a/b - 7

=> 2 (root 10) = a - 7b / b

=> root 10 = a - 7b / 2b

Since,a and b are integers therefore, a-7b, 2b are also integers . We have b=0 therefore , 2b=0.

root 10 is rational no. which is a contradiction because root is irrational therefore , our assumption is wrong .

So, 7+2 root 5 irrational no.

HOPE YOU UNDERSTOOD THIS

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