Question

Rationalisation of denominator
$\frac{4}{3 \sqrt{3} - 2 \sqrt{2} } + \frac{3}{3 \sqrt{3} + 2 \sqrt{2} }$
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Answer 1

$: \longrightarrow{ \underline{{ \huge{ \mathfrak{R}}} \mathfrak{equired} \: \: \: { \huge{ \mathfrak{A}}} \mathfrak{nswer :-}}}$

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$\tt{ \: 1) \: \frac{4}{3 \sqrt{3} - 2 \sqrt{2} } + \frac{3}{3 \sqrt{3} + 2 \sqrt{2} } }$

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Now, take the LCM,

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We know that,

(a+b) (a-b) = a² - b²

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$: \longrightarrow { \tt{ \: \frac{4(3 \sqrt{3} + 2 \sqrt{2}) + 3(3 \sqrt{3} - 2 \sqrt{2}) }{( {3 \sqrt{3}) ^{2} - (2 \sqrt{2} )}^{2} } }}$

We know that,

√3 × √3 = 3

$: \longrightarrow{ \tt{ \: \frac{(12\sqrt{3}) - ( {8 \sqrt{2} ) + 9√3 - 2√2} }{27 - 8} }}$

$: \longrightarrow{ \tt{ \: \frac{7(27 - 8)}{19} }}$

$: \longrightarrow{ \tt{ \: \frac{21√3 - 10√2}{ \cancel{19}} }}$

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You can also substitute it,

We know that,

√3 = 1.732 ; √2 = 1.414

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$: \longrightarrow{ \tt{ \: \frac{36.372 - 14.14}{ \cancel{19}} }}$

$: \longrightarrow{ \tt{ \: \frac{22.232}{ \cancel{19}} }}$

$\boxed{ \blue{ : \longrightarrow { \tt{ \:1.17\: \: ..ans }}}}$

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@MissSolitary ✌️

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Answer 2

Solution!!

$\sf \dfrac{4}{3\sqrt{3}-2\sqrt{2}}+\dfrac{3}{3 \sqrt{3}+2\sqrt{2}}$

Taking the LCM

$\sf = \dfrac{4(3\sqrt{3}+2\sqrt{2})+3(3\sqrt{3}-2\sqrt{2})}{(3\sqrt{3}-2\sqrt{2})(3\sqrt{3}+2\sqrt{2})}$

Rationalising the denominator using (a + b)(a - b) = a² - b²

$\sf = \dfrac{4(3\sqrt{3}+2\sqrt{2})+3(3\sqrt{3}-2\sqrt{2})}{(3\sqrt{3})^{2}-(2\sqrt{2})^{2}}$

$\sf = \dfrac{4(3\sqrt{3}+2\sqrt{2})+3(3\sqrt{3}-2\sqrt{2})}{9(3)-4(2)}$

$\sf = \dfrac{4(3\sqrt{3}+2\sqrt{2})+3(3\sqrt{3}-2\sqrt{2})}{19}$

Distribute 4 and 3 through the parentheses

$\sf = \dfrac{12\sqrt{3}+8\sqrt{2}+9\sqrt{3}-6\sqrt{2}}{19}$

Collecting the like term

$\boxed{\sf = \dfrac{21\sqrt{3}+2\sqrt{2}}{19}}$

Rationalising the denominator:-

In brief, it is the process by which a fraction is simplified so that the denominator is a rational number.

More identities:-

→ (a + b)² = a² + b² + 2ab

→ (a - b)² = a² + b² - 2ab

→ (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

→ (a + b)³ = a³ + b³ + 3ab(a + b)

→ (a - b)³ = a³ - b³ - 3ab(a - b)

→ (a + b)(a - b) = a² - b²