If x-\frac{2}{x}=15, then what is the value of (x^{2}+\frac{4}{x^{2}}) ?

SSC CGL 24 Aug 2021 Shift 2 Paper

Question : 52 of 100

Marks: +1, -0

x-\;{2}/{x}=15

, then what is the value of

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

Solution:

Given,

x-\;{2}/{x}=15

Square on both sides we get,

⇒ \;\; (x-\;{2}/{x})^{2}=(15)^{2}

⇒ x^{2}+\;{4}/{x^{2}}-2 × x × \;{2}/{x}=225

[ ∵ (a-b)^{2}=a^{2}+b^{2}-2 a b]

⇒ \;\; x^{2}+\;{4}/{x^{2}}=225+4

∴ \;\; x^{2}+\;{4}/{x^{2}}=229

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