rationalise root 3- root 4/root 3 +root 4- root 7 - Maths - Polynomials

Question

rationalise root 3- root 4/root 3 +root 4- root 7

Ankush Jain · Accepted Answer

We have,

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

On rationalising the denominator, we get

$\frac{\sqrt{3} - 2}{(\sqrt{3} + 2) - \sqrt{7}} = \frac{\sqrt{3} - 2}{(\sqrt{3} + 2) - (\sqrt{7})} \times \frac{(\sqrt{3} + 2 + \sqrt{7})}{(\sqrt{3} + 2) + (\sqrt{7})} = \frac{(\sqrt{3} - 2) (\sqrt{3} + 2 + \sqrt{7})}{(\sqrt{3} + 2)^{2} - 7} = \frac{(\sqrt{3} - 2) (\sqrt{3} + 2 + \sqrt{7})}{3 + 4 + 4 \sqrt{3} - 7} = \frac{(\sqrt{3} - 2) (\sqrt{3} + 2 + \sqrt{7})}{4 \sqrt{3}} = \frac{(\sqrt{3}) (\sqrt{3} - 2) (\sqrt{3} + 2 + \sqrt{7})}{12}$