If \sqrt{x}-\frac{1}\sqrt{x}=\sqrt{7}, then the value of x^{2}+\frac{1}{x^{2}} is

SSC CGL 20 Aug 2021 Shift 2 Paper

Question : 69 of 100

Marks: +1, -0

√{x}-\;{1}/{√{x}}=√{7}

, then the value of

x^{2}+\;{1}/{x^{2}}

Solution:

Given,

√{x}-\;{1}/{√{x}}=√{7}

Square on both sides,

(√{x}-\;{1}/{√{x}})^{2}=(√{7})^{2}

⇒(√{x})^{2}+(\;{1}/{√{x}})^{2}-2 × √{x} × \;{1}/{√{x}}=7

⇒ \;\; x+\;{1}/{x}=7+2=9

Again, square both side,

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

⇒\;x^{2}+\;{1}/{x^{2}}+2 × x × \;{1}/{x}=81

⇒\;x^{2}+\;{1}/{x^{2}}=81-2=79

∴ \;x^{2}+\;{1}/{x^{2}}=79

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