The number which exceeds its positive Square roots by 12 is:
Correct Answer: Description for Correct answer:
Let the required number is x
According to given condition
\( \Large x=\sqrt{x}+12 => x-12=\sqrt{x} \)
=> \( \Large x^{2}-25x+144=0 \)
=> \( \Large x^{2}-16x-9x+144=0 \)
Therefore, x = 16, 9
since x = 9 does not hold the condition
Therefore, x = 16
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