Question

The Value of $\mathtt{\sqrt {3-2 \sqrt 2}}$ is

(a) $\mathtt{\sqrt 3 + \sqrt 2}$
(b) $\mathtt{\sqrt 3 - \sqrt 2}$
(c) $\mathtt{\sqrt 2 + 1}$
(d) $\mathtt{\sqrt 2 - 1}$​

Answer 1

$\large{\frak \green{To \: Solve :}}$

$\mathtt{\sqrt {3-2 \sqrt 2}}$

$\large{ \frak \red{Solution :}}$

$\tt{ \implies\mathtt{\sqrt {3-2 \sqrt 2}} }$

$\mathtt{ \implies\sqrt{(2^2) - 2 \sqrt{2} \times 1 + {1}^{2} } }$

$\implies{ \mathtt{ \sqrt{ \sqrt{2} + 1 }^2 }}$

$\Huge{ \implies\pink{ \boxed{\underline{\underline{ \mathtt{ \sqrt{2} + 1 }}}}}}$

Answer 2

Answer:

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Step-by-step explanation:

$\begin{gathered} \sqrt{3 - 2 \sqrt{2} } \\ \\ \sqrt{2 - 2 \sqrt{2} + 1} \\ \\ = \sqrt{ { (\sqrt{2} )}^{2} - (2 \times 1 \times \sqrt{2}) + {1^2 }} \\ \\ = \sqrt{ {(\sqrt{2} - 1) }^{2} } \\ \\ = \sqrt{2} - 1 \end{gathered}$

∴ Option d