SSC CPO SI and ASI 2016 Paper

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Question : 142 of 200

Marks: +1, -0

If

x = 1 + √2 + √3

and

y = 1 + √2 – √3

, then the value of

{x^{2}+4 x y+y^{2}}/{x+y}

is

6
$2(1 + √2)$
$1$
$2√2$

Solution:

(1) Let

a=1+√{2}

and

b=√{3},

so,

x=a+b

and

y=a-b

. Now

{x^{2}+4 x y+y^{2}}/{x+y}

=\ {(a+b)^{2}+4 (a^{2}-b^{2} )+(a-b)^{2}}/{a+b+a-b}

= {a^{2}+b^{2}+2 a b+4 a^{2}-4 b^{2}+a^{2}+b^{2}-2 a b}/{2 a}

= {3 a^{2}-b^{2}}/{a}

= {3+6+6 √{2}-3}/{1+√{2}}=6

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