If $x^{2} + y^{2} = 25, x y = 12$ , then $x, y =$

(3, 4)

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(3, - 3)

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(3, 4)

(- 3, - 4)

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(- 3, - 3)

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Solution

The correct option is C $(3, 4)$ , $(- 3, - 4)$
$x^{2} + y^{2} = 25$ and
$x y = 12 \Rightarrow y = \frac{12}{x}$
$∴$ Replace $y$ in first equation $\Rightarrow$
$x^{2} + {(\frac{12}{x})}^{2} = 25$
$\Rightarrow x^{4} + 144 = 25 x^{2}$
$\Rightarrow x^{4} - 25 x^{2} + 144 = 0$
$\Rightarrow x^{4} - 16 x^{2} - 9 x^{2} + 144 = 0$
$\Rightarrow x^{2} (x^{2} - 16) - 9 (x^{2} - 16) = 0$
$\Rightarrow (x^{2} - 9) (x^{2} - 16) = 0$
$\Rightarrow x^{2} = 9$ or $x^{2} = 16$
$\Rightarrow x = \pm 3$ or $x = \pm 4$
$∴ c$

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