To find the set of values of $k$ for which the curve $y = kx^2 + 2x - k$ and the line $y = kx - 2$ do not intersect, we need to determine when these two equations have no common solutions.

Step 1: Set the equations equal to each other

Since we want to find where they do not intersect, let's set the two equations equal to each other to find where they might intersect:

$kx^2 + 2x - k = kx - 2$

Step 2: Rearrange the equation

Rearrange this equation to set it equal to zero:

$kx^2 + 2x - k - kx + 2 = 0$

Combine like terms:

$kx^2 + (2 - k)x + (2 - k) = 0$

Step 3: Analyze the quadratic equation

Now, we have a quadratic equation:

$kx^2 + (2 - k)x + (2 - k) = 0$

For the curve and line to not intersect, this quadratic equation should have no real solutions. This happens when the discriminant is less than zero.

Step 4: Calculate the discriminant

The discriminant $\Delta$ of a quadratic equation $ax^2 + bx + c = 0$ is given by:

$\Delta = b^2 - 4ac$

In our equation, $a = k$, $b = 2 - k$, and $c = 2 - k$. Substituting these into the discriminant formula, we get:

$\Delta = (2 - k)^2 - 4(k)(2 - k)$

Step 5: Set up the inequality

For the quadratic to have no real solutions, the discriminant must be less than zero:

$(2 - k)^2 - 4k(2 - k) < 0$

Step 6: Expand and solve the inequality

First, expand both terms:

$(2 - k)^2 = 4 - 4k + k^2$

$4k(2 - k) = 8k - 4k^2$

Substituting these into the inequality:

$4 - 4k + k^2 - (8k - 4k^2) < 0$

Combine like terms:

$4 - 4k + k^2 - 8k + 4k^2 < 0$

$5k^2 - 12k + 4 < 0$

Step 7: Solve the quadratic inequality

To solve $5k^2 - 12k + 4 < 0$, we first find the roots of the equation:

$5k^2 - 12k + 4 = 0$

Using the quadratic formula $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$k = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \cdot 5 \cdot 4}}{2 \cdot 5}$

$k = \frac{12 \pm \sqrt{144 - 80}}{10}$

$k = \frac{12 \pm \sqrt{64}}{10}$

$k = \frac{12 \pm 8}{10}$

$k = \frac{20}{10} \text{ or } \frac{4}{10}$

$k = 2 \text{ or } k = 0.4$

Step 8: Determine the intervals

The quadratic $5k^2 - 12k + 4$ is a parabola opening upwards. Thus, it is less than zero between its roots:

$0.4 < k < 2$

Conclusion

The set of values of $k$ for which the curve $y = kx^2 + 2x - k$ and the line $y = kx - 2$ do not intersect is:

$k \in (0.4, 2)$

Would you like to see more details or have any other questions?

Related Questions:

1. What happens when $k = 0.4$ or $k = 2$?
2. How can we find the points of intersection for other values of $k$?
3. How would the solution change if the line equation were different?
4. What is the geometric interpretation of the discriminant in this context?
5. How do we analyze the case where the curve and line intersect at exactly one point?

Tip: Always double-check your discriminant calculations when working with quadratic equations to ensure accurate results.