Question

rationalise 4 upon root 9 - root 1 + root 3 be
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Answer 1

Step-by-step explanation:

$\frac{4}{ \sqrt{9 } - \sqrt{1} + \sqrt{3} } \\ = \frac{4}{ 3 - 1 + \sqrt{3} } \\ = \frac{4}{2 + \sqrt{3} } \\ = \frac{4}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ = \frac{8 - 4 \sqrt{3} }{4 - 3} \\ = \frac{8 - 4 \sqrt{3} }{1} \\ = 8 - 4 \sqrt{3}$