Question

On rationalising the denominator of 1/root3-root2

Answer 1

Answer:

$\sqrt{3} + \sqrt{2}$

Step-by-step explanation:

1/$\sqrt{3} - \sqrt{2}$ *$\sqrt{3} + \sqrt{2}$/$\sqrt{3} + \sqrt{2}$ =

$\sqrt{3} + \sqrt{2}$/3-2 =

$\sqrt{3} + \sqrt{2}$/1 =

$\sqrt{3} + \sqrt{2}$

Answer 2

1/√3-√2

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

Remember:

a²-b²=(a+b)(a-b)