Given that $f(x) = 3x^2 + 8x - 35$, where $x \in \mathbb{R}$, find the two roots of $f(x) = 0$ - Leaving Cert Mathematics - Question b - 2021

Rearranging the equation gives:

$3^{2m + 1} + 8(3^m) - 35 = 0$

Using the substitution $n = 3^m$, we transform the equation to:

$3n^2 + 8n - 35 = 0$

Now applying the quadratic formula:

$n = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 3 \cdot (-35)}}{2 \cdot 3}$

Calculating the discriminant:

$64 + 420 = 484$

Then:

$n = \frac{-8 \pm 22}{6}$

This leads to:

1. $n = \frac{14}{6} = \frac{7}{3}$
2. $n = \frac{-30}{6} = -5$ (not valid since $n = 3^m > 0$)

Thus, we have:

$3^m = \frac{7}{3}$

Taking logarithms yields:

$m = \log_3 \left(\frac{7}{3}\right)$

This can be expressed as:

$m = \log_3 7 - \log_3 3 = \log_3 7 - 1$

So our answer is $m = \log_3 7 - 1$, which is in the form $m = \log_b p - q$ where $p = 7$, $q = 1$, and $b = 3$.