Question

Prove that 3+2√7 is irrational number

Answer 1

I hope this will help you

Answer 2

$SOLUTION:- \:$

Let ,


$x = 3 + 2 \sqrt{7}$


Squaring both the sides, we get:-



$= > x {}^{2} = (3 + 2 \sqrt{7} ) {}^{2}$




$= > x^{2} = 3^{2} + 2 \times 3 \times 2 \sqrt{7} + (2 \sqrt{7} ) {}^{2}$




$= > x^{2} = 9 + 12 \sqrt{7} + 4 \times 7$





$= > x^{2} = 9 + 12 \sqrt{7} + 28$





$= > x^{2} = 37 + 12 \sqrt{7}$





$= >x {}^{2} - 37 = 12 \sqrt{7}$





$= > \frac{x {}^{2} - 37 }{12} = \sqrt{7}$




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$Therefore, \:$




$\sqrt{7} \: is \: an \: irrational \: number.$




HENCE PROVED