NIMCET 2009 Question Paper with solutions

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Question : 23 of 120

Marks: +1, -0

If

2 x^{4}+x^{3}-11 x^{2}+x+2=0

, then the values of

x+\;{1}/{x}

are

$-3, \;{5}/{2}$
$-\;{5}/{2}, 3$
$\;{2}/{5}, \;{1}/{3}$
$\;{1}/{3},-5$

Solution:

2 x^{4}+x^{3}-11 x^{2}+x+2=0

⇒ 2 x^{2}+x-11+\;{1}/{x}+\;{2}/{x^{2}}=0

(dividing by

x^{2}

)

=2(x^{2}+\;{1}/{x^{2}})+(x+\;{1}/{x})-11=0

=2[(x+\;{1}/{x})^{2}-2]+(x+\;{1}/{x})-11=0

On putting

x+\;{1}/{x}=y

, we get

2 y^{2}-2+y-11=0

⇒ \;\; 2 y^{2}+y-15=0

⇒ \;\; y=\;{-1 ± √{1+120}}/{4}=\;{-1 ± 11}/{4}

⇒ \;\; y=\;{5}/{2},-3 ⇒ x+\;{1}/{x}=-3, \;{5}/{2}

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