The sum of a natural number and its positive square root is 132. Find the number.

Let the required natural number be x.

According to the given condition,

$x+\sqrt{x}=132$

Putting $\sqrt{x}=y$ or $x=y^{2}$, we get

$y^{2}+y=132$

$\Rightarrow y^{2}+y-132=0$

$\Rightarrow y^{2}+12 y-11 y-132=0$

$\Rightarrow y(y+12)-11(y+12)=0$

$\Rightarrow(y+12)(y-11)=0$

$\Rightarrow y+12=0$ or $y-11=0$

$\Rightarrow y=-12$ or $y=11$

∴ y = 11         (y cannot be negative)

Now,

$\sqrt{x}=11$

$\Rightarrow x=(11)^{2}=121$

Hence, the required natural number is 121.