Question

1/root3-2
rationa;ise the denominator of the following

Answer 1

Step-by-step explanation:

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

Answer 2

Step-by-step explanation:

$\frac{1}{ \sqrt{3} - 2 }$

Now in order to rationalise the denominator we first multiply opposite pair i.e., √3+2 in numerator and denominator, so we get:

[tex] \frac{1}{ \sqrt{3 } - 2} \times \frac{ \sqrt{3} + 2 }{ \sqrt{3} + 2} [/tex]

= -√3 + 2