Find the range of values of $k$ for which $kx^2 + 8x + k < 6$ for all real values of $x$. Given the curve $y = 8x^2 + mx + m - 6$, where $m$ is a constant, find the possible values of $m$ for which the x-axis is a tangent to the curve. the range of values of $m$ for which the curve has a positive y-intercept. Show that the equation $2x^2 - cx + c^2 + 6 = 0$ has no real roots for all real values of $c$.

1. Rearrange the inequality into standard quadratic form and state the conditions for it to be always negative

$kx^2 + 8x + k < 6$
$kx^2 + 8x + (k - 6) < 0$
For the quadratic expression to be always negative for all real values of $x$, we must have:

1. Coefficient of $x^2 < 0 \implies k < 0$
2. Discriminant $< 0$

2. Set up and simplify the discriminant inequality

$b^2 - 4ac < 0$
$(8)^2 - 4(k)(k - 6) < 0$
$64 - 4k^2 + 24k < 0$
Divide by $-4$ (and flip the inequality sign):
$k^2 - 6k - 16 > 0$

3. Solve the quadratic inequality for $k$

$(k - 8)(k + 2) > 0$
This gives the ranges:
$k < -2 \quad \text{or} \quad k > 8$

4. Combine the conditions to find the final range of $k$

Since we established earlier that $k < 0$, we must reject $k > 8$.
Taking the intersection of $k < 0$ and $(k < -2 \text{ or } k > 8)$:
$k < -2$

1. State the condition for the x-axis to be a tangent to the curve

For the x-axis ($y = 0$) to be a tangent to the curve, the equation $8x^2 + mx + m - 6 = 0$ must have two equal real roots, so the discriminant $b^2 - 4ac = 0$.

2. Form the equation in terms of $m$

$m^2 - 4(8)(m - 6) = 0$
$m^2 - 32m + 192 = 0$

3. Solve the quadratic equation for $m$

$(m - 8)(m - 24) = 0$
$m = 8 \text{ or } m = 24$

1. Find the expression for the y-intercept

When $x = 0$,
$y = 8(0)^2 + m(0) + m - 6$
$y = m - 6$

2. Set up and solve the inequality for a positive y-intercept

For a positive y-intercept, $y > 0$:
$m - 6 > 0$
$m > 6$

1. Formulate the discriminant of the quadratic equation in $x$

For the equation $2x^2 - cx + (c^2 + 6) = 0$, the discriminant is:
$\Delta = b^2 - 4ac$
$\Delta = (-c)^2 - 4(2)(c^2 + 6)$

2. Simplify the discriminant and show it is always negative

$\Delta = c^2 - 8c^2 - 48$
$\Delta = -7c^2 - 48$
Since $c^2 \geqslant 0$ for all real values of $c$, $-7c^2 \leqslant 0$.
Therefore, $-7c^2 - 48 \leqslant -48 < 0$.
Since $\Delta < 0$ for all real $c$, the equation has no real roots.