Question

please do it fast and give me the ans



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Answer 1

ANSWER

your answer is 2.4142135624

Answer 2

$\sqrt{3 + 2 \sqrt{2} }$

You can Do It by 2 methods.

1st method

$let \: \sqrt{ 3 + 2 \sqrt{2} } = \sqrt{x } + \sqrt{y}$

Squaring Both Sides.

$( \sqrt{3 + 2\sqrt{2} } ) {}^{2} = ( \sqrt{x} + \sqrt{y} ) {}^{2} \\ \\ 3 + 2 \sqrt{2} = x + y + 2 \sqrt{xy}$

On Comparing Both sides we got.

$x + y = 3 \\ \\ 2 \sqrt{xy} = 2 \sqrt{2} \\ \\ \sqrt{xy} = \frac{2 \sqrt{2} }{2} \\ \\ \sqrt{xy} = \sqrt{2} \\ \\ xy = \sqrt{2} {}^{2} \\ \\ xy = 2$

So X can be=2

Y can be =1

As we know

$\sqrt{3 + 2 \sqrt{2} } = \sqrt{x} + \sqrt{y} \\ \\ \sqrt{3 + 2 \sqrt{2} } = \sqrt{2} + \sqrt{1} \\ \\ \sqrt{3 + 2 \sqrt{2} } = \sqrt{2} + 1$

2nd method

As we know (a+b)²=a²+b²+2ab

Here Written

$3 + 2 \sqrt{2} \\ \\ 1 + 2 + 2 \sqrt{2} \\ \\ ( \sqrt{2} + 1) {}^{2}$

By breaking the. terms we get =(√2+1)²

$\sqrt{3 + 2 \sqrt{2} } \\ \\ \sqrt{( \sqrt{2} + 1) {}^{2} } \\ \\ (\sqrt{2} + 1) {}^{2 \times \frac{1}{2} } \\ \\ = \sqrt{2} + 1$

$\underline{ \huge{ \sqrt{3 + 2 \sqrt{2} } = \sqrt{2} + 1}}$