If x+ \frac{1}{x}=3, where x \neq 0, then the value of {x^{4}+3 x^{3}+5 x^{2}+3 x+1}/{x^{4}+1} is

SSC CPO SI and ASI 2016 Paper

Question : 116 of 200

Marks: +1, -0

x+\ {1}/{x}=3,

where

x \ ≠ 0,

then the value of

\ {x^{4}+3 x^{3}+5 x^{2}+3 x+1}/{x^{4}+1}

Solution:

(3)

x+\ {1}/{x}=3 \ ⇒\ (x+\ {1}/{x}\ )^{2}=9

\ ⇒ x^{2}+\ {1}/{x^{2}}=7

Now,

\ {x^{4}+3 x^{3}+5 x^{2}+3 x+1}/{x^{4}+1}

=\ {\ (x^{2}+ {1}/{x^{2}}\ )+3\ (x+\ {1}/{x}\ )+5}/{x^{2}+ {1}/{x^{2}}}

=\ {7+9+5}/{7}=\ {21}/{7}=3

Go to Question:

Prev Question

Next Question