Question

prove that 3 + 2 root 2 the whole square is irrational​

Answer 1

Step-by-step explanation:

$(a + b) {}^{2} = a {}^{2} + 2ab + b {}^{2 } \\ = ( 3 + 2 \sqrt{2} ) {}^{2} \\ = 3 { + (2 \sqrt{2)} }^{2} + 2 \times 3 \times 2 \sqrt{2} \\ = 9 + 4 \times 2 + 12 \sqrt{2} \\ = 26 + 12 \sqrt{2}$

$hence \: \sqrt{2} is \: an \: irrational \: \\ number$

$so \: 26 + 12 \sqrt{2} is \: also \: an \\ irrational \: number$

$26 + 12 \sqrt{2} = 3 + 2 \sqrt{2}$

$therefore \: \\ 3 + 2 \sqrt{2} \: is \: an \: irrational \: number$

Producet of rational and irrational number is irrational number.

Answer 2

Step-by-step explanation:

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