MCQ
The number which exceeds its positive square root by $12$ is
- A$9$
- ✓$16$
- C$25$
- DNone of these
So, $x = \sqrt x + 12\,\,\, \Rightarrow x - 12 = \sqrt x $
==> ${x^2} - 25x + 144 = 0$
==> ${x^2} - 16x - 9x + 144 = 0$$ \Rightarrow x = 16$
=Since $x = 9$ does not hold the condition.
Trick : By inspection, since $16$ exceeds its positive square root i.e.,$4$ by $12$.
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