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SSC CGL 18 Aug 2021 Shift 2 Paper

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Question : 54 of 100

Marks: +1, -0

If

x^{4}+y^{4}+x^{2} y^{2}=21

x^{2}+y^{2}-x y=7

, then what is the value of

\;\frac{x}{y}+\;\frac{y}{x}

?

$\;\frac{5}{4}$
$\;\frac{3}{4}$
$-\;\frac{3}{2}$
$-\;\frac{5}{2}$

Solution:

Given

x^{4}+y^{4}+x^{2} y^{2}=21

and

\;\; x^{2}+y^{2}-x y=7

We can write

x^{4}+y^{4}+2 x^{2} y^{2}-x^{2} y^{2}\;=21

\Rightarrow \;\; (x^{2}+y^{2})^{2}-x^{2} y^{2}=21

. . . (i)

\; [ \because (a+b)^{2}=a^{2}+b^{2}+2ab ]

. . . (ii)

Put in Eq. (i) we get,

(7+x y)^{2}-x^{2} y^{2} =21 \ldots \;\text{ (iii) }\;

\Rightarrow 49+x^{2} y^{2}+14 x y-x^{2} y^{2} \;=21

\Rightarrow \; 14 x y=21-49 \; =-28

x y \; =-2

\therefore \left(\;\frac{x}{y}+\;\frac{y}{x}\right) \; =\; \frac{x^{2}+y^{2}}{x y}

\; =\; \frac{7+x y}{x y} \; \;\text{[By Eq. (ii)]}\;

\; =\; \frac{7-2}{-2} = -\; \frac{5}{2}

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