Question

1/4+3 root5 rationalize the denominator

Answer 1

To be Rationalized:

$\Longrightarrow \sf \dfrac{1}{4 + 3\sqrt{5}}$

Solution:

$\Longrightarrow \sf \dfrac{1}{4+3\sqrt{5}}\\ \\ \\ \\\Longrightarrow \sf \dfrac{1}{4+3\sqrt{5}} \times \dfrac{4-3\sqrt{5}}{4-3\sqrt{5}}\\ \\ \\ \\\Longrightarrow \sf \dfrac{4-3\sqrt{5}}{(4+3\sqrt{5}) (4-3\sqrt{5})}\\ \\ \\ \\Using \ a^2 - b^2 = (a + b)(a - b);\\ \\ \\ \Longrightarrow \sf \dfrac{4-3\sqrt{5}}{(4)^2-(3\sqrt{5})^2}\\ \\ \\ \\\Longrightarrow \sf \dfrac{4-3\sqrt{5}}{16-9(5)}\\ \\ \\ \\\Longrightarrow \sf \dfrac{4-3\sqrt{5}}{16-45}\\ \\ \\ \\\Longrightarrow \sf \dfrac{4-3\sqrt{5}}{-29 \ }$

Hence rationalized.

Answer 2

ᗩᘉSᘺᘿᖇ

$\frac{1}{4 + 3 \sqrt{5} } \times \frac{4 - 3 \sqrt{5} }{4 - 3 \sqrt{5} }$

↣$\frac{4 - 3 \sqrt{5} }{{(4) }^{2} - ( {3 \sqrt{5}) }^{2} }$

↣$\frac{4 - 3 \sqrt{5} }{16 - (</h3><h3>9 \times 5)}$

↣$\frac{4 - 3 \sqrt{5} }{16 - 45}$

↣$\frac{4-3\sqrt{5}}{-29}$

↣$\frac{-4+3\sqrt{5}}{29}$