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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.

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