The number which exceeds its positive square | vedclass

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(B) Let the required number be $x$. According to the problem,$x = \sqrt{x} + 12$. Rearranging the equation,we get $x - 12 = \sqrt{x}$. Squaring both s

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(B) Let the required number be $x$. According to the problem,$x = \sqrt{x} + 12$. Rearranging the equation,we get $x - 12 = \sqrt{x}$. Squaring both sides,we get $(x - 12)^2 = x$. $x^2 - 24x + 144 = x$. $x^2 - 25x + 144 = 0$. Factoring the quadratic equation: $(x - 16)(x - 9) = 0$. So,$x = 16$ or $x = 9$. Check for $x = 16$: $16 - \sqrt{16} = 16 - 4 = 12$. This satisfies the condition. Check for $x = 9$: $9 - \sqrt{9} = 9 - 3 = 6 
eq 12$. This does not satisfy the condition. Therefore,the correct number is $16$.

The number which exceeds its positive square root by $12$ is

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