Question

please solve 9 and 10 no.

Answer 1

$\bf{\large{\underline{\underline{Question:-}}}}$

9. Simplify $\sf{ \sqrt{3 + 2 \sqrt{2}}}$

10. Simplify $\sf{ \sqrt{3-2 \sqrt{2}}}$

$\bf{\large{\underline{\underline{Answer:-}}}}$

$\boxed{\sf{9.\: \sqrt{3 + 2 \sqrt{2}} = \sqrt{2} +1}}$

$\boxed{10.\:\sf{ \sqrt{3 - 2 \sqrt{2}} = \sqrt{2} -1}}$

$\bf{\large{\underline{\underline{Explanation:-}}}}$

9. Simplify $\bf{ \sqrt{3 + 2 \sqrt{2}}}$

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

It can be written as :

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

It can be written as,

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

Now it is in the form of x² + y² + 2xy

We know that (x + y)² = x² + y² + 2xy

Here x = √2 and y = 1

By substituting the values in the identity we have,

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

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

$\boxed{\sf{ \sqrt{3 + 2 \sqrt{2}} = \sqrt{2} +1}}$

10. Simplify $\bf{ \sqrt{3 - 2 \sqrt{2}}}$

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

It can be written as :

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

It can be written as,

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

Now it is in the form of x² + y² - 2xy

We know that (x - y)² = x² + y² - 2xy

Here x = √2 and y = 1

By substituting the values in the identity we have,

$\tt{= \sqrt{ {( \sqrt{2} - 1)}^{2} }}$

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

$\boxed{\sf{ \sqrt{3 - 2 \sqrt{2}} = \sqrt{2} -1}}$

$\bf{\large{\underline{\underline{Identity\:Used:-}}}}$

[1] (x + y)² = x² + y² + 2xy

[2] (x - y)² = x² + y² - 2xy

$\bf{\large{\underline{\underline{Some\:Important\:Identities:-}}}}$

[1] (x + y)² = x² + y² + 2xy

[2] (x - y)² = x² + y² - 2xy

[3] (x + y)(x - y) = x² - y²

[4] (x + a)(x + b) = x² + (a + b)x + ab