SSC CHSL 17 Mar 2020 Shift 3 Paper

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Question : 51 of 100

Marks: +1, -0

If

x=1+\√{2}

, then find the value of

\√{x}+({1}/{\√{x}})

$2.1014$
$2.1973$
$1.9876$
$1.9996$

Solution:

Given,

x=1+\√{2}

...(i)
Then,

{1}/{x}={1}/{1+\√{2}}

⇒

{1}/{x}={1}/{1+\√{2}} ×{1-\√{2}}/{1-\√{2}}

⇒

{1}/{x}={1-\√{2}}/{1-2}=\√{2}-1

...(ii)
Now,

(\√{x}+{1}/{\√{x}})^{2}=x+{1}/{x}+2

⇒

(\√{x}+{1}/{\√{x}})^{2}=1+\√{2}+\√{2}-1+2

[from Eqs. (i) and (ii)]
⇒

(\√{x}+{1}/{\√{x}})^{2}=2+2 \√{2}

⇒

(\√{x}+{1}/{\√{x}})^{2}=2+2.828

⇒

(\√{x}+{1}/{\√{x}})^{2}=4.828

⇒

\√{x}+{1}/{\√{x}}=\√{4.828}

=2.1973

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