Question

Please send the solution

Answer 1

Heya mate!!

$x = 3 + 2 \sqrt{2} \\ its \: square \: root \\ \sqrt{3 + 2 \sqrt{2} } \\ \\ \sqrt{ {1 + \sqrt{2} }^{2} } \\ 1 + \sqrt{2}$

$x = 1 + \sqrt{2} \\ \frac{1}{x} = \frac{1}{1 + \sqrt{2} } \\ \frac{1}{1 + \sqrt{2} } \times \frac{1 - \sqrt{2} }{1 - \sqrt{2} } \\ \frac{1 - \sqrt{2} }{ - 1} = - 1 + \sqrt{2}$

$x - \frac{1}{x} \\ 1 + \sqrt{2} - ( - 1 + \sqrt{2} ) \\ 1 + \sqrt{2} + 1 - \sqrt{2} \\ 2$

Hope it helps u dear✌️

Answer 2

GIVEN:

★$x =3+2\sqrt{2}$

TO FIND:

★$\sqrt{x}-\dfrac{1}{\sqrt{x}}$

ANSWER:

We have,

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

On putting square root both sides, we have,

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

=>$\sqrt{x}=\sqrt{2+1+2\sqrt{2}}$

=>$\sqrt{x}=\sqrt{\sqrt{(2)^{2}}+1+2\sqrt{2}}$

=>$\sqrt{x}=\sqrt{\sqrt{(2)^{2}}+1+2×1×\sqrt{2}}$

$\large\red{\boxed{(a+b)^{2}=a^{2}+b^{2}+2ab}}$

=>$\sqrt{x} = \sqrt{(\sqrt{2}+1) ^{2}}$

=>$\sqrt{x}=\sqrt{2}+1$

$\large\green{\boxed{\sqrt{x}=\sqrt{2}+1}}$

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Now,

$\dfrac{1}{\sqrt{x}}=\dfrac{1}{\sqrt{2}+1}$

On rationalizing the denominator,

=>$\dfrac{1}{\sqrt{x}}=\dfrac{1\times(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}$

=>$\dfrac{1}{\sqrt{x}}=\dfrac{(\sqrt{2}-1)}{(\sqrt{2})^{2}-(1)^{2}}$

$\large\purple{\boxed{(a+b)(a-b)=a^{2}-b^{2}}}$

=>$\dfrac{1}{\sqrt{x}}=\dfrac{(\sqrt{2}-1)}{2-1}$

=>$\dfrac{1}{\sqrt{x}}=\dfrac{(\sqrt{2}-1)}{1}$

$\large\blue{\boxed{\dfrac{1}{\sqrt{x}}=(\sqrt{2}-1)}}$

_____________________________________

Now,

$\sqrt{x}-\dfrac{1}{\sqrt{x}}$

=$\sqrt{2}+1-(\sqrt{2}-1)$

=$\sqrt{2}+1-\sqrt{2}+1$

=$2$

Hence , $\sqrt{x}-\dfrac{1}{\sqrt{x}}=2$

$\huge\orange{\boxed{.°.\sqrt{x}-\dfrac{1}{\sqrt{x}}=2}}$