MCQ
If ${x^2} + {y^2} = 25,\;xy = 12$, then $x = $
- A$\{3, 4\}$
- B$\{3, -3\}$
- ✓$\{3, 4, -3, -4\}$
- D$\{-3, -3\}$
==> ${x^2} + {\left( {\frac{{12}}{x}} \right)^2} = 25\,\,$
$\Rightarrow {x^4} + 144 - 25{x^2} = 0$
==> $({x^2} - 16)({x^2} - 9) = 0$
==> ${x^2} = 16$ and ${x^2} = 9$
==> $x = \pm \,4$ and $x = \pm 3$.
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