Show that the roots of the equation $ax^2 + (3a+b)x + 3b = 0$ are real for all real values of $a$ and $b$. Find the range of values of $m$ for which the line $y = mx - 3$ will never cut the curve $y^2 = 4x - 6y - 34$.

1. Write down the expression for the discriminant of the quadratic equation.

For the equation $ax^2 + (3a+b)x + 3b = 0$, the discriminant is:
$\Delta = (3a+b)^2 - 4(a)(3b)$

2. Expand and simplify the discriminant.

$\Delta = 9a^2 + 6ab + b^2 - 12ab$
$\Delta = 9a^2 - 6ab + b^2$

3. Factorise the discriminant into a perfect square and conclude.

$\Delta = (3a - b)^2$
Since $(3a - b)^2 \ge 0$ for all real values of $a$ and $b$, $\Delta \ge 0$.
Therefore, the roots of the equation are always real.

1. Substitute the equation of the line into the equation of the curve

Substitute $y = mx - 3$ into $y^2 = 4x - 6y - 34$:
$(mx - 3)^2 = 4x - 6(mx - 3) - 34$

2. Expand and simplify to form a quadratic equation in $x$

$m^2x^2 - 6mx + 9 = 4x - 6mx + 18 - 34$
$m^2x^2 - 6mx + 9 = 4x - 6mx - 16$
$m^2x^2 - 4x + 25 = 0$

3. Apply the discriminant condition for no intersection

For the line to never cut the curve, the quadratic equation must have no real roots, so discriminant $< 0$:
$b^2 - 4ac < 0$
$(-4)^2 - 4(m^2)(25) < 0$
$16 - 100m^2 < 0$

4. Solve the quadratic inequality for $m$

$100m^2 > 16$
$m^2 > \frac{16}{100}$
$m^2 > \frac{4}{25}$
$m < -\frac{2}{5} \text{ or } m > \frac{2}{5}$