Question

rationalize the denominator 7/3 √3-2 √2

Answer 1

Given:

$\sf\to\dfrac{7}{3 \sqrt{3} - 2 \sqrt{2} }$

Find:

➝ Rationalise the denominator.

Solution:

$\sf\to\dfrac{7}{3 \sqrt{3} - 2 \sqrt{2}}$

$\sf := \dfrac{7}{3 \sqrt{3} - 2 \sqrt{2}} \times \dfrac{3 \sqrt{3} + 2 \sqrt{2} }{3 \sqrt{2} + 2 \sqrt{2} }$

using identity :-

(a+b) (a-b) = a² - b²

So,

$\sf := \dfrac{7(3 \sqrt{3} + 2 \sqrt{2})}{ {(3 \sqrt{3} )}^{2} - {(2 \sqrt{2})}^{2} }$

$\sf := \dfrac{7 \times 3 \sqrt{3} + 7 \times 2 \sqrt{2}}{(9 \times 3) - (4 \times 2)}$

$\sf := \dfrac{21\sqrt{3} + 14\sqrt{2}}{(9 \times 3) - (4 \times 2)}$

$\sf := \dfrac{21\sqrt{3} + 14\sqrt{2}}{(27 - 8)}$

$\sf := \dfrac{21\sqrt{3} + 14\sqrt{2}}{19}$

Hence, the rationalized denomiator is 19

Answer 2

GIVEN :-

• 7/(3√3 - 2√2).

TO FIND :-

• The rationalize form of 7/(3√3 - 2√2).

SOLUTION :-

☯ $\\ \underline{\boldsymbol{According\: to \:the\: Question\:now :}}$

$: \implies \displaystyle \sf \: \dfrac{7}{3 \sqrt{3} - 2 \sqrt{2} }$

$: \implies \displaystyle \sf \: \dfrac{7}{3 \sqrt{3} - 2 \sqrt{2} } \times \frac{3 \sqrt{3} + 2 \sqrt{2} }{3 \sqrt{3} + 2 \sqrt{2} }$

$: \implies \displaystyle \sf \: \dfrac{7 \bigg(3 \sqrt{3} + 2 \sqrt{2 } \bigg) }{ \bigg(3 \sqrt{3} - 2 \sqrt{2} \bigg) \bigg(3 \sqrt{3} + 2 \sqrt{2} \bigg) }$

$\dag \displaystyle \rm \: using \: identity : (a + b)(a - b) = a^{2} - b^{2}$

$: \implies \displaystyle \sf \: \frac{21 \sqrt{3} + 14 \sqrt{2} }{ \bigg(3 \sqrt{3} \bigg) ^{2} - \bigg(2 \sqrt{2} \bigg) ^{2} }$

$: \implies \displaystyle \sf \: \frac{21 \sqrt{3} + 14 \sqrt{2} }{9 \sqrt{9} - 4 \sqrt{4} }$

$: \implies \displaystyle \sf \: \frac{21 \sqrt{3} + 14 \sqrt{2} }{9 \times 3 - 4 \times 2}$

$: \implies \displaystyle \sf \: \frac{21 \sqrt{3} + 14 \sqrt{2} }{27 - 8}$

$: \implies \displaystyle \sf \: \frac{21 \sqrt{3} + 14 \sqrt{2} }{19}$

$\therefore \underline {\displaystyle \sf required \ answer \ is \ \frac{21 \sqrt{3} + 14 \sqrt{2} }{19}}$