Question

If √2 = 1.414 and √ 3 = 1.732 then find 4/3√3-2√2 +3/3√3+2√2

Answer 1

Hey friend,
Here is the answer you were looking for:
$\frac{4}{3 \sqrt{3} - 2 \sqrt{2} } + \frac{3}{3 \sqrt{3} + 2 \sqrt{2} } \\ \\ on \: rationalizing \: the \: denominator \: we \: get \\ \\ = \frac{4}{3 \sqrt{3} - 2 \sqrt{2} } \times \frac{3 \sqrt{3} + 2 \sqrt{2} }{3 \sqrt{3} + 2 \sqrt{2} } + \frac{3}{3 \sqrt{3} + 2 \sqrt{2} } \times \frac{3 \sqrt{3} - 2 \sqrt{2} }{3 \sqrt{3} - 2 \sqrt{2} } \\ \\ using \: the \: identity \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{4 \times 3 \sqrt{3} + 4 \times 2 \sqrt{2} }{ {(3 \sqrt{3}) }^{2} - {(2 \sqrt{2}) }^{2} } + \frac{3 \times 3 \sqrt{3} - 3 \times 2 \sqrt{2} }{ {(3 \sqrt{3}) }^{2} - {(2 \sqrt{2}) }^{2} } \\ \\ = \frac{12 \sqrt{3} + 8 \sqrt{2} }{27 - 8} + \frac{9 \sqrt{3} - 6 \sqrt{2} }{27 - 8} \\ \\ = \frac{12 \sqrt{3} + 8 \sqrt{2} + 9 \sqrt{3} - 6 \sqrt{2} }{19} \\ \\ = \frac{21 \sqrt{3} + 2 \sqrt{2} }{19} \\ \\ = \frac{21 \times 1.732 + 2 \times 1.414}{19} \\ \\ = \frac{36.372 + 2.828}{19} \\ \\ = \frac{39.2}{19}$

Hope this helps!!!!

@Mahak24

Thanks...
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