Question

rationalize the denominator​

Answer 1

$\texttt{\textsf{\large{\underline{Solution}:}}}$

Given fraction,

$\tt = \dfrac{ \sqrt{3} + 2 }{ \sqrt{3} - 2 }$

Multiplying both numerator and denominator by (√3 + 2), we get,

$\tt = \dfrac{( \sqrt{3} + 2 )^{2} }{( \sqrt{3} - 2 )( \sqrt{3} + 2)}$

Using identity (a + b)(a - b) = a² - b², we get,

$\tt = \dfrac{( \sqrt{3} + 2 )^{2} }{( \sqrt{3})^{2} - (2)^{2} }$

$\tt = \dfrac{( \sqrt{3} + 2 )^{2} }{3 - 4}$

$\tt = \dfrac{( \sqrt{3} + 2 )^{2} }{ - 1}$

$\tt = - ( \sqrt{3} + 2 )^{2}$

$\tt = - \{ ( \sqrt{3})^{2} +( 2 )^{2} + 2 \times 2 \times \sqrt{3} \}$

$\tt = - \{3 +4 + 4 \sqrt{3}\}$

$\tt = -(7 + 4 \sqrt{3})$

$\tt = -7 - 4 \sqrt{3}$

which is our required answer.

$\texttt{\textsf{\large{\underline{Answer}:}}}$

• -7 - 4√3.

•••♪

Answer 2

Step-by-step explanation:

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