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The sum of a natural number and its positive square root is 132. The number is [[1]].

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answer is 121.

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Detailed Solution

The sum of a natural number and its positive square root is 132. The number is 121.
Given that the sum of the natural number and its positive square root is 132.
Let’s assume that the two natural number is

m

.
According to the given condition,

⟹ m + \sqrt{m} = 132

Substitute

\sqrt{m} = n

⟹ n^{2} + n = 132

⟹ n^{2} + n - 132 = 0

⟹ n^{2} + 12 n - 11 n - 132 = 0

⟹ n (n + 12) - 11 (n + 12) = 0

⟹ (n - 11) (n + 12) = 0

⟹ n - 11 = 0 or n + 12 = 0

∴

n = 11 or n = - 12

Here, -12 is not a natural number
So, we will choose the value

n = 11

,
And

⟹ \sqrt{m} = n

⟹ \sqrt{m} = 11

By squaring on the both sides, we get,

∴ m = 121

Thus, required natural number is 121

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