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, $a + b + n = 3 - 2 + 5 = 6$.
If the set of all values of \( a \), for which the equation \( 5x^3 - 15x - a = 0 \) has three distinct real roots, is the interval \( (\alpha, \beta) \), then \( \beta - 2\alpha \) is equal to
If the equation \( a(b - c)x^2 + b(c - a)x + c(a - b) = 0 \) has equal roots, where \( a + c = 15 \) and \( b = \frac{36}{5} \), then \( a^2 + c^2 \) is equal to .