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### Problem Breakdown:

We are given:

- The equation of a curve \( C \) is:
  \[
  2x^2 + y^2 + 4x - 2y = 3
  \]
  
- The equation of a line \( l \) is:
  \[
  y = 2x + k
  \]
  where \( k \) is a constant.

#### Part (a): 
We are tasked to show that the \( x \)-coordinates of the points where \( C \) and \( l \) intersect are given by the solutions to the equation:
\[
6x^2 + 4kx + k^2 - 2k - 3 = 0
\]

#### Part (b): 
We are to find the two values of \( k \) for which \( l \) is a tangent to \( C \).

---

### Solution:

#### Part (a): 
We need to substitute \( y = 2x + k \) into the equation of the curve and simplify.

The curve's equation is:
\[
2x^2 + y^2 + 4x - 2y = 3
\]
Substitute \( y = 2x + k \):

\[
2x^2 + (2x + k)^2 + 4x - 2(2x + k) = 3
\]

Now, expand and simplify:
\[
2x^2 + (4x^2 + 4kx + k^2) + 4x - (4x + 2k) = 3
\]

\[
2x^2 + 4x^2 + 4kx + k^2 + 4x - 4x - 2k = 3
\]

Simplify further:
\[
6x^2 + 4kx + k^2 - 2k = 3
\]

Move the 3 to the other side:
\[
6x^2 + 4kx + k^2 - 2k - 3 = 0
\]

This is the required equation, which proves part (a).

---

#### Part (b):
For the line \( l \) to be tangent to the curve \( C \), the quadratic equation:
\[
6x^2 + 4kx + k^2 - 2k - 3 = 0
\]
must have exactly one solution, meaning the discriminant \( \Delta \) of the quadratic equation should be zero.

The general form of the quadratic equation is:
\[
ax^2 + bx + c = 0
\]
where:
- \( a = 6 \)
- \( b = 4k \)
- \( c = k^2 - 2k - 3 \)

The discriminant \( \Delta \) is given by:
\[
\Delta = b^2 - 4ac
\]

Substitute \( a = 6 \), \( b = 4k \), and \( c = k^2 - 2k - 3 \):
\[
\Delta = (4k)^2 - 4(6)(k^2 - 2k - 3)
\]

Simplify:
\[
\Delta = 16k^2 - 24(k^2 - 2k - 3)
\]

Distribute the -24:
\[
\Delta = 16k^2 - 24k^2 + 48k + 72
\]

Simplify:
\[
\Delta = -8k^2 + 48k + 72
\]

Set \( \Delta = 0 \) for tangency:
\[
-8k^2 + 48k + 72 = 0
\]

Divide the entire equation by -8:
\[
k^2 - 6k - 9 = 0
\]

Solve this quadratic equation using the quadratic formula:
\[
k = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-9)}}{2(1)}
\]

Simplify:
\[
k = \frac{6 \pm \sqrt{36 + 36}}{2}
\]
\[
k = \frac{6 \pm \sqrt{72}}{2}
\]
\[
k = \frac{6 \pm 6\sqrt{2}}{2}
\]
\[
k = 3 \pm 3\sqrt{2}
\]

Thus, the two values of \( k \) are:
\[
k = 3 + 3\sqrt{2} \quad \text{and} \quad k = 3 - 3\sqrt{2}
\]

These are the two values of \( k \) for which the line is tangent to the curve.

---

Do you want more details or have any questions about the solution? 

Here are 5 related questions you might be interested in:
1. How do we verify the discriminant condition for tangency?
2. What happens when the discriminant is greater than zero in this context?
3. Can we graphically represent the curve and tangent lines for specific \( k \) values?
4. How would the solution change if the curve equation were different?
5. What if the line equation had a different slope, like \( y = 3x + k \)?

**Tip:** Always check the discriminant of a quadratic equation when determining the number of solutions (intersections).

The curve C has equation 2x^2 + y^2 + 4x - 2y = 3. The line l has equation y = 2x + k, where k is a constant. (a) Show that the x-coordinates of the points where C and l intersect are given by the solutions to the equation 6x^2 + 4kx + k^2 - 2k - 3 = 0. (b) Hence find the two values of k for which l is a tangent to C.

Solve a problem involving a curve and a line intersection by finding the quadratic equation for the x-coordinates of intersection points and determining the values of k where the line is tangent to the curve. Learn about discriminants and quadratic equations with this step-by-step solution.

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