Question

7/√2-√6 rationalize the denominator? ​

Answer 1

${ \color{red} \hookrightarrow}\bf Question$

Rationalise the denominator :-

$\dfrac{7}{ \sqrt{2} - \sqrt{6} }$

Solution

$: \implies \dfrac{7}{ \sqrt{2} - \sqrt{6} } \times \dfrac{\sqrt{2} + \sqrt{6} }{\sqrt{2} + \sqrt{6} }$

In denominator , following algebraic identity can be used

$\rightarrow \rm(a - b)(a + b) = \bf{a}^{2} - {b}^{2}$

On Implementation,

$: \implies \dfrac{7 \sqrt{2} + 7 \sqrt{6} }{ (\sqrt{2}) {}^{2} - {( \sqrt{6}) }^{2} }$

$\rightarrow \dfrac{7 (\sqrt{2} + \sqrt{6})}{2 - 6} = \dfrac{7( \sqrt{2} + \sqrt{6})}{ - 4}$

Since negative sign in denominator doesn't exist so it would be

$:\implies \pink{\dfrac{ - 7( \sqrt{2} + \sqrt{6})}{ 4} } \red{ \ast}$

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More to know

1. In order to rationalise an irrational number having surd simply multiply the numerator and denominator with the conjugates of denominator.
2. Surds are irrational so we can't write them in the form of fraction or whole number.
3. The term ' surd ' is different from the term 'radical '
4. Never leave your answer in surd form.

Thankyou