Express each one of the following with rational denominator: (i) 1/3 + √2 (ii) 1/√6 - √5 (iii) 16/√41 - 5

Question

Express each one of the following with rational denominator:

(i) \(\frac{1}{3+\sqrt{2}}\)

(ii) \(\frac{1}{\sqrt6-\sqrt{5}}\)

(iii) \(\frac{16}{\sqrt41-5}\)

Answer 1

(i) we have \(\frac{1}{3+\sqrt{2}}\) rationalizing factor of the denominator is 3 -√2

= \(\frac{1}{3+\sqrt{2}}\) = \(\frac{1}{3+\sqrt{2}}\) x \(\frac{3-\sqrt2}{3-\sqrt{2}}\)

= \(\frac{3-\sqrt2}{(3)^2-(\sqrt{2})^2}\)

because (a + b)(a - b) = (a)² - (b)²

= \(\frac{3-\sqrt2}{9-2}\) = \(\frac{3-\sqrt2}{7}\)

(ii) we have \(\frac{1}{\sqrt6-\sqrt{5}}\) rationalizing factor of the denominator is √6 + √5

= \(\frac{1}{\sqrt6-\sqrt{5}}\) x \(\frac{\sqrt6+\sqrt5}{\sqrt6+\sqrt{5}}\)

= \(\frac{\sqrt6+\sqrt5}{(\sqrt6)^2-(\sqrt{5})2}\) = \(\frac{\sqrt6+\sqrt{5}}{6-5}\) = \(\frac{\sqrt6+\sqrt{5}}{1}\)

= √6 + √5

(iii) we have \(\frac{16}{\sqrt41-5}\) rationalizing factor of the denominator is √41 + 5

= \(\frac{16}{\sqrt41-5}\) x \(\frac{\sqrt41+5}{\sqrt41+5}\)

= \(\frac{16\times (\sqrt41+5)}{(\sqrt41-5)(\sqrt41+5)}\) = \(\frac{16\sqrt41+5}{(\sqrt41)^2-(5)^2}\)

= \(\frac{16\sqrt41-5}{41-25}\) = \(\frac{16\sqrt41+5}{16}\) = √41 + 5