If x= {1}/{2+\sqrt{3}, y= {1}/{2-\\sqrt{3}, then the value of 8x y (x^{2}+y^{2} ) is

SSC CPO SI and ASI 2015 Paper

Question : 125 of 199

Marks: +1, -0

x= {1}/{2+√{3}}, y= {1}/{2-\√{3}}, \

then the value of 8

x y\ (x^{2}+y^{2}\ )

Solution:

(3)
Here

x= {1}/{2+\√{3}} ×{2-\√{3}}/{2-\√{3}}

= {2-√{3}}/{4-3}=2-\√{3}

And

y= {1}/{2-\√{3}} \×{2+\√{3}}/{2+\√{3}}

=\ {2+\√{3}}/{4-3}=2+√{3}

Therefore

x+y=2-\√{3}+2+\√{3}=4

and

x y=4-3=1

Now,
8

x y\ (x^{2}+y^{2}\ )

=8 \× x y\ ((x+y)^{2}-2 x y\ )

8 \×(4^{2}-2 \× 1\ )=8 \×(16-2)=112

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