Question

Rationalise 7 root 3-5 root 2 / root 48 + root18

Answer 1

Hey friend,
Here is the answer you were looking for:
$\frac{7 \sqrt{3} - 5 \sqrt{2} }{ \sqrt{48} + \sqrt{18} } \\ \\ = \frac{7 \sqrt{3} - 5 \sqrt{2} }{ \sqrt{2 \times 2 \times 2 \times 2 \times 3} + \sqrt{2 \times 3 \times 3} } \\ \\ = \frac{7 \sqrt{3} - 5 \sqrt{2} }{2 \times 2 \sqrt{3} + 3 \sqrt{2} } \\ \\ = \frac{7 \sqrt{3} - 5 \sqrt{2} }{4 \sqrt{3} + 3 \sqrt{2} } \\ \\ on \: rationalizing \: the \: denominator \: we \: get \\ \\ = \frac{7 \sqrt{3} - 5 \sqrt{2} }{4 \sqrt{3} + 3 \sqrt{2} } \times \frac{4 \sqrt{3} - 3 \sqrt{2} }{4 \sqrt{3} - 3 \sqrt{2} } \\ \\ using \: the \: identity \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{7 \sqrt{3} \times 4 \sqrt{3} - 7 \sqrt{3} \times 3 \sqrt{2} - 5 \sqrt{2} \times 4 \sqrt{3} + 5 \sqrt{2} \times 3 \sqrt{2} }{ {(4 \sqrt{3}) }^{2} - {(3 \sqrt{2}) }^{2} } \\ \\ = \frac{28 \times 3 - 21 \sqrt{6} - 20 \sqrt{6} + 15 \times 2 }{48 - 18} \\ \\ = \frac{84 - 41\sqrt{6} + 30}{30} \\ \\ = \frac{114 - 41 \sqrt{6} }{30}$


Hope this helps!!!

@Mahak24

Thanks...
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Answer 2

Answer:

$\frac{114-41\sqrt{6}}{30}$

Step-by-step explanation:

Given expression is,

$\frac{7\sqrt{3}-5\sqrt{2}}{\sqrt{48}+\sqrt{18}}$

For rationalizing,

Multiply √48 - √18 on both numerator and denominator,

$=\frac{7\sqrt{3}-5\sqrt{2}}{\sqrt{48}+\sqrt{18}}\times \frac{\sqrt{48}-\sqrt{18}}{\sqrt{48}-\sqrt{18}}$

$=\frac{7\sqrt{144}-5\sqrt{96}-7\sqrt{54}+5\sqrt{36}}{(\sqrt{48})^2-(\sqrt{18})^2}$

$=\frac{7\times 12-5\times 4\sqrt{6}-7\times 3\sqrt{6}+5\times 6}{48-18}$

$=\frac{84-20\sqrt{6}-21\sqrt{6}+30}{30}$

$=\frac{114-41\sqrt{6}}{30}$