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Let's solve the questions step by step.

### 1. (a) The quadratic equation is:

\[
kx^2 + 4x + (5 - k) = 0
\]

This is a quadratic equation in the form \( ax^2 + bx + c = 0 \), where:

- \( a = k \)
- \( b = 4 \)
- \( c = 5 - k \)

For the quadratic to have **two different real solutions**, the discriminant must be greater than 0. The discriminant \( \Delta \) for a quadratic equation \( ax^2 + bx + c = 0 \) is given by:

\[
\Delta = b^2 - 4ac
\]

Substitute \( a = k \), \( b = 4 \), and \( c = 5 - k \):

\[
\Delta = 4^2 - 4(k)(5 - k)
\]

Simplifying:

\[
\Delta = 16 - 4[k(5 - k)] = 16 - 4(5k - k^2)
\]
\[
\Delta = 16 - 20k + 4k^2
\]

To ensure two different real solutions, \( \Delta > 0 \):

\[
4k^2 - 20k + 16 > 0
\]

Divide through by 4:

\[
k^2 - 5k + 4 > 0
\]

This proves part (a), where \( k \) satisfies \( k^2 - 5k + 4 > 0 \).

### 1. (b) Finding the set of possible values of \( k \):

We now need to solve the inequality \( k^2 - 5k + 4 > 0 \). First, solve the equation \( k^2 - 5k + 4 = 0 \) to find the critical points.

Using the quadratic formula:

\[
k = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(4)}}{2(1)}
\]
\[
k = \frac{5 \pm \sqrt{25 - 16}}{2}
\]
\[
k = \frac{5 \pm \sqrt{9}}{2}
\]
\[
k = \frac{5 \pm 3}{2}
\]

So, the two solutions are:

\[
k = \frac{5 + 3}{2} = 4 \quad \text{and} \quad k = \frac{5 - 3}{2} = 1
\]

Thus, \( k^2 - 5k + 4 = 0 \) has roots at \( k = 1 \) and \( k = 4 \). To solve \( k^2 - 5k + 4 > 0 \), consider the sign of the quadratic expression in intervals around the roots \( k = 1 \) and \( k = 4 \):

- For \( k < 1 \), \( k^2 - 5k + 4 > 0 \)
- For \( 1 < k < 4 \), \( k^2 - 5k + 4 < 0 \)
- For \( k > 4 \), \( k^2 - 5k + 4 > 0 \)

Thus, the inequality \( k^2 - 5k + 4 > 0 \) holds for:

\[
k \in (-\infty, 1) \cup (4, \infty)
\]

### 2. (a) Completing the square:

#### For \( x^2 - 6x + 27 \):

Start with \( x^2 - 6x + 27 \). To complete the square:

\[
x^2 - 6x = (x - 3)^2 - 9
\]

So,

\[
x^2 - 6x + 27 = (x - 3)^2 - 9 + 27 = (x - 3)^2 + 18
\]

#### For \( 1 + 6x - x^2 \):

Rearrange \( 1 + 6x - x^2 \) as \( -x^2 + 6x + 1 \), then factor out \( -1 \):

\[
- (x^2 - 6x - 1) = -( (x - 3)^2 - 9 - 1) = -( (x - 3)^2 - 10)
\]

So,

\[
1 + 6x - x^2 = - (x - 3)^2 + 10
\]

### 2. (b) Sketching the curves:

You are asked to sketch:

1. \( y = x^2 - 6x + 27 \), which is \( y = (x - 3)^2 + 18 \)
2. \( y = 1 + 6x - x^2 \), which is \( y = - (x - 3)^2 + 10 \)

Both curves are parabolas with their vertices at \( x = 3 \). The first parabola opens upwards with vertex \( (3, 18) \), and the second opens downwards with vertex \( (3, 10) \).

The shortest distance between the two curves is the vertical distance between their vertices:

\[
\text{Shortest distance} = 18 - 10 = 8
\]

This completes the solution.

Would you like more details or have any specific questions?

### Related Questions:
1. How do you derive the quadratic formula?
2. What are the conditions for a quadratic equation to have no real solutions?
3. How do you complete the square for a general quadratic expression?
4. How do you sketch parabolas from the completed square form?
5. How can the discriminant help determine the number of roots in a cubic equation?

**Tip**: Completing the square is a useful technique not only for solving quadratics but also for analyzing the graph of a quadratic function.

1. The equation kx^2 + 4x + (5 - k) = 0 has 2 different real solutions for x. Find the values of k. 2. Express quadratic equations in completed square form and find shortest distance between curves.

This problem involves solving quadratic equations with real roots by analyzing the discriminant, expressing quadratic functions in completed square form, and finding the shortest distance between parabolas. A step-by-step guide is provided, suitable for Grade 10-12 students.

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