Question

Rationalise the denominator of the following.
√3+2/√3-2​

Answer 1

To rationalize the denominator,

$\frac{ \sqrt{3} + 2 }{ \sqrt{3} - 2 } \\ \\ = \frac{ \sqrt{3} + 2 }{ \sqrt{3} - 2 } \times \frac{ \sqrt{3} + 2 }{ \sqrt{3} + 2 } \\ \\ = \frac{( \sqrt{3} + 2) {}^{2} }{ \sqrt{3 {}^{2} } - 2{}^{2} } \\ \\ = \frac{7 + 4 \sqrt{3} }{ - 1} \\ \\ = - (7 + 4 \sqrt{3})$

Hence,the rationalized form is -(7+4√3)

What is rationalising?

Rationalising is a mathematical expression in which the denominator of a fraction is multiplied or divided to remove the radical signs.

But why do we rationalize?

For some mathematical procedures,radical signs in the denominator are not required and make the solution complicated,to avoid this we rationalize

Answer 2

$\large{\underline{\boxed{\tt Answer}}}$

$-(7+4\sqrt{3}$

$\large{\underline{\boxed{\tt Explanation}}}$

Givem Problem:

Rationalise the denominator of the following.

√3+2/√3-2

Solution:

To Rationalise:

Denominator of √3+2/√3-2

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Method:

⇒$\frac{\sqrt{3+} 2}{\sqrt{3-} 2}$

⇒$=\frac{\sqrt{3} +2}{\sqrt{3-} 2} * \frac{\sqrt{3+} 2}{\sqrt{3+} 2}$

⇒$\frac{(\sqrt{3+2})^2 }{\sqrt{3^2-} 2^2}$

⇒$\frac{7+4\sqrt{3} }{-1}$

⇒$-(7+4\sqrt{3})$

Hence,

The rationalized form is -(7+4√3)