Question

Rationalise 10/√7-√2 ​

Answer 1

$\frac{10}{\sqrt{7} } -\sqrt{2} =\frac{10\sqrt{7} -7\sqrt{2} }{7}$

( Decimal: $2.36543...$)

Steps:

1: Convert element to fraction.

$\sqrt{2} =\frac{\sqrt{2}\sqrt{7} }{\sqrt{7} }$

$=\frac{10}{\sqrt{7} } -\frac{\sqrt{2} \sqrt{7} }{\sqrt{7} }$

2: Since denominator are equal, combine the fractions.

$\frac{a}{c}$ ± $\frac{b}{c} =\frac{a+b}{c}$

$=\frac{10-\sqrt{2} \sqrt{7} }{\sqrt{7} }$

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$\sqrt{2} \sqrt{7} =\sqrt{14}$

$=\frac{10-\sqrt{14} }{\sqrt{7} }$

Rationalize $\frac{10-\sqrt{14} }{\sqrt{7} }$ : $\frac{10\sqrt{7} -7\sqrt{2} }{7}$

$=\frac{10\sqrt{7}-7\sqrt{2} }7}$

Answer 2

Solution:

$\frac{10}{ \sqrt{7} - \sqrt{2} } \\ = \frac{10}{ \sqrt{7} - \sqrt{2} } \times \frac{ \sqrt{7} + \sqrt{2} }{ \sqrt{7} + \sqrt{2} } \\ = \frac{10( \sqrt{7} + \sqrt{2}) }{( \sqrt{7} - \sqrt{2} )( \sqrt{7} + \sqrt{2} )} \\ = \frac{10 \sqrt{7} + 10 \sqrt{2} }{( { \sqrt{7} )}^{2} - ( \sqrt{2} ) ^{2} } \\ = \frac{10( \sqrt{7} + \sqrt{2} )}{7 - 2} \\ = \frac{10( \sqrt{7} + \sqrt{2} ) }{5} \\ = 2( \sqrt{7} + \sqrt{2} ) \\ = 2 \sqrt{7} + 2 \sqrt{2}$