Question

prove (3√2+√5) is an rational number and also check whether (3√2+√5)(3√2-√5)  is irrational or rational

Answer 1

let x be the number that is a rational number equal to the value of 3√2+√5.
Now,
x= 3√2+√5.
x²=(3√2+√5)²
x²=18+5+6√10
x²-23=6√10
x²-23÷6=√10.
now,
x is a rational no.
⇒x² is a rational number.
⇒x²-23÷6 is a rational no.
⇒√10 is a rational no.
But, √10 is a irrational no.
we arrive contradiction. thus  3√2+√5 is a irrational no.

( 3√2+√5)( 3√2-√5)⇒(a+b)(a-b)
⇒a²-b²
⇒ (3√2)²-(√5)
⇒9×2-5
⇒13
It is a rational no.

Answer 2

let x = (3√2 + √5)
squarring on both sides
x² = (3√2 + √5)²
x² = 9×2 + 5 +6√10
x²- 23 = 6√10
x² - 23 ÷ 6 =√10
x² - 23 ÷ 6 is a rational number
therefore √10 is also rational
but √10 is irrational
therefore this is a contradiction
therefore (3√2 + √5) is an irrational number

(3√2+√5)(3√2-√5) = a+b * a-b = a²-b²
(3√2)²-(√5)²
9×2 - 5
18 - 5
13 
it is a rational number