If x = 3 - sqrt {5,} then {{sqrt x } over {s | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(d) $x = 3 - \sqrt 5 $

$\sqrt x = \sqrt {3 - \sqrt 5 } = {1 \over {\sqrt 2 }}\,.\sqrt {6 - 2\sqrt 5 } = {1 \over {\sqrt 2 }}(\sqrt 5 - 1)$  &n

Vedclass · Accepted Answer

$1/\sqrt 5 $

Vedclass · Answer

(d) $x = 3 - \sqrt 5 $

$\sqrt x = \sqrt {3 - \sqrt 5 } = {1 \over {\sqrt 2 }}\,.\sqrt {6 - 2\sqrt 5 } = {1 \over {\sqrt 2 }}(\sqrt 5 - 1)$         $3x - 2 = 9 - 3\sqrt 5 - 2 = 7 - 3\sqrt 5 = {{14 - 6\sqrt 5 } \over 2}$

= ${{{{(3 - \sqrt 5 )}^2}} \over 2}$;

$ \Rightarrow $ $\sqrt 2 + \sqrt {3x - 2} = \sqrt 5 \,.\,\sqrt x $;

$	herefore {{\sqrt x } \over {\sqrt 2 + \sqrt {3x - 2} }} = {1 \over {\sqrt 5 }}$.

If $x = 3 - \sqrt {5,} $ then ${{\sqrt x } \over {\sqrt 2 + \sqrt {(3x - 2)} }} = $

Similar Questions

The square root of $\frac{(0.75)^3}{1-(0.75)}+\left[0.75+(0.75)^2+1\right]$ is

If $a = \sqrt {(21)} - \sqrt {(20)} $ and $b = \sqrt {(18)} - \sqrt {(17),} $ then

The square root of $\sqrt {(50)} + \sqrt {(48)} $ is

The number of integers $q , 1 \leq q \leq 2021$, such that $\sqrt{ q }$ is rational, and $\frac{1}{ q }$ has a terminating decimal expansion, is

$\root 4 \of {(17 + 12\sqrt 2 )} = $