SSC CGL 23 Aug 2021 Shift 1 Paper

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Question : 60 of 100

Marks: +1, -0

If

x+\;{1}/{x}=\;{17}/{4}, x > 1

, then what is the value of

x-\;{1}/{x} ?

$\;{8}/{3}$
$\;{15}/{4}$
$\;{3}/{2}$
$\;{9}/{4}$

Solution:

Given,

(x+\;{1}/{x})=\;{17}/{4}

Square on both sides,

(x+\;{1}/{x})^{2}=(\;{17}/{4})^{2}

⇒ \;\; x^{2}+\;{1}/{x^{2}} +2 × x × \;{1}/{x}=\;{289}/{16}

⇒ \;\; x^{2}+\;{1}/{x^{2}} \;\; =\;{289}/{16}-2

\;\; =\;{289-32}/{16}=\;{257}/{16} . . . \;\text " (i) "\;

Now,

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

⇒ \;\; (x-\;{1}/{x})^{2}= \;\; \;{257}/{16}-2

\{ ∵ x^{2}+\;{1}/{x^{2}}=\;{257}/{16}\}

\;\; =\;{(257-32)}/{16}

⇒ \;\; (x-\;{1}/{x})^{2} \;\; =\;{225}/{16}

⇒ \;\; (x-\;{1}/{x}) \;\; =√{\;{225}/{16}}

(x-\;{1}/{x}) \;\; =\;{15}/{4}

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