Question

rationalise the denominator √3 + 1
-----------
√3 - 1​

Answer 1

$\;\large{{\sf{\purple{★彡Answer彡★ }}}}$

Identities used :

$\begin{gathered} {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \end{gathered}$

Now,

$\begin{gathered} \frac{ \sqrt{3} + 1}{ \sqrt{3} - 1 } \\ \end{gathered}$

On rationalizing the denominator we get

Now,

$\begin{gathered} \frac{ \sqrt{3} + 1}{ \sqrt{3} - 1 } \\ \end{gathered}$

On rationalizing the denominator we get,

$\begin{gathered} = \frac{ \sqrt{3} + 1 }{ \sqrt{3} - 1} \times \frac{ \sqrt{3} + 1 }{ \sqrt{3} + 1 } \\ \\ = \frac{ {( \sqrt{3}) }^{2} + {(1)}^{2} + 2( \sqrt{3})(1) }{ {( \sqrt{3}) }^{2} - {(1)}^{2} } \\ \\ = \frac{3 + 1 + 2 \sqrt{3} }{3 - 1} \\ \\ = \frac{4 + 2 \sqrt{3} }{2} \\ \\ = \frac{2(2 + \sqrt{3}) }{2} \\ \\ = 2 + \sqrt{3} \end{gathered}$

Answer 2

Answer:

$\;\large{{\sf{\purple{★彡Answer彡★ }}}}$

hope this helps you ⤴️⤴️