Question

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 ?

• A. 4
• B. 22
• C. 18
• D. 6

Answer 1

Answer: (C)

18

Answer 2

We are given that:

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

Squaring both sides:

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

Equating the rational and irrational parts:
- $a^2 + b^2n = 29$ (rational part)
- $2ab\sqrt{n} = -12\sqrt{5}$ (irrational part)

From $2ab\sqrt{n} = -12\sqrt{5}$, comparing the terms under the square root gives $n = 5$, so:

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

Now, using $a^2 + b^2n = 29$, we substitute $n = 5$:

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

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

By trial and error or systematic solving, we find $a = 3$, $b = -2$, and $n = 5$.
Thus, $a + b + n = 3 - 2 + 5 = 6$.