Question:medium

If $(a + b\sqrt{n})$ is the positive square root of $(29 - 12\sqrt{5})$, where $a$ and $b$ are integers, and $n$ is a natural number, then the maximum possible value of $(a + b + n)$ is ?

Updated On: Jan 13, 2026
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The Correct Option is C

Solution and Explanation

We start with the given equation:

$\sqrt{29 - 12\sqrt{5}} = a + b\sqrt{n}$

Square both sides of the equation:

$29 - 12\sqrt{5} = (a + b\sqrt{n})^2 = a^2 + 2ab\sqrt{n} + b^2n$

By comparing the rational and irrational parts of the equation, we get two separate equations:
- $a^2 + b^2n = 29$ (for the rational parts)
- $2ab\sqrt{n} = -12\sqrt{5}$ (for the irrational parts)

From the irrational part, $2ab\sqrt{n} = -12\sqrt{5}$, we can deduce that $n = 5$. Substituting this into the equation yields:

$2ab\sqrt{5} = -12\sqrt{5} \implies ab = -6$

Now, substitute $n = 5$ into the rational part equation, $a^2 + b^2n = 29$:

$a^2 + 5b^2 = 29$

We now have a system of two equations:
1. $ab = -6$
2. $a^2 + 5b^2 = 29$

Solving these equations (either by trial and error or systematically) gives us the values $a = 3$, $b = -2$, and $n = 5$.
Therefore, the sum $a + b + n = 3 + (-2) + 5 = 6$.

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