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The problem involves a quadratic equation with the condition that it has a double root.

### Given:
The quadratic equation is:
\[
x^2 + (k - 4)x + 2k - 11 = 0
\]
and the condition says the equation has one double real root.

### Step 1: Condition for a double root
A quadratic equation has a double root if the **discriminant** is equal to zero. The general form of a quadratic equation is \( ax^2 + bx + c = 0 \), and the discriminant \( \Delta \) is given by:
\[
\Delta = b^2 - 4ac
\]
For the equation to have a double root, we require:
\[
\Delta = 0
\]

### Step 2: Identify coefficients
From the given equation \( x^2 + (k - 4)x + (2k - 11) = 0 \), we identify the coefficients:
- \( a = 1 \)
- \( b = k - 4 \)
- \( c = 2k - 11 \)

### Step 3: Set discriminant to zero
Using the discriminant formula:
\[
\Delta = (k - 4)^2 - 4(1)(2k - 11)
\]
Set this equal to zero for the double root condition:
\[
(k - 4)^2 - 4(2k - 11) = 0
\]
Now we solve for \( k \).

1. Expand \( (k - 4)^2 \):
\[
(k - 4)^2 = k^2 - 8k + 16
\]
2. Expand \( 4(2k - 11) \):
\[
4(2k - 11) = 8k - 44
\]
3. Now, substitute these into the discriminant equation:
\[
k^2 - 8k + 16 - (8k - 44) = 0
\]
Simplify:
\[
k^2 - 8k + 16 - 8k + 44 = 0
\]
\[
k^2 - 16k + 60 = 0
\]

### Step 4: Solve the quadratic equation for \( k \)
We now solve \( k^2 - 16k + 60 = 0 \) using the quadratic formula:
\[
k = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(1)(60)}}{2(1)}
\]
\[
k = \frac{16 \pm \sqrt{256 - 240}}{2}
\]
\[
k = \frac{16 \pm \sqrt{16}}{2}
\]
\[
k = \frac{16 \pm 4}{2}
\]
This gives two solutions for \( k \):
\[
k_1 = \frac{16 + 4}{2} = 10, \quad k_2 = \frac{16 - 4}{2} = 6
\]

### Step 5: Solve the quadratic equation for each \( k \)
Now, substitute both values of \( k \) into the original equation to solve for the double root.

#### For \( k = 10 \):
The equation becomes:
\[
x^2 + (10 - 4)x + 2(10) - 11 = 0
\]
\[
x^2 + 6x + 9 = 0
\]
This simplifies to:
\[
(x + 3)^2 = 0
\]
Thus, the double root is \( x = -3 \).

#### For \( k = 6 \):
The equation becomes:
\[
x^2 + (6 - 4)x + 2(6) - 11 = 0
\]
\[
x^2 + 2x + 1 = 0
\]
This simplifies to:
\[
(x + 1)^2 = 0
\]
Thus, the double root is \( x = -1 \).

### Final Answer:
(a) The values of \( k \) are \( k = 10 \) and \( k = 6 \).

(b) For \( k = 10 \), the double root is \( x = -3 \). For \( k = 6 \), the double root is \( x = -1 \).

Let me know if you would like further details on any step.

### Related Questions:
1. What is the general process to find double roots for any quadratic equation?
2. How is the discriminant related to the nature of roots in a quadratic equation?
3. Can a quadratic equation have more than one double root?
4. What happens when the discriminant is positive or negative?
5. How do the solutions of a quadratic equation change as the value of \( k \) varies?

### Tip:
Always check if your discriminant calculations simplify correctly, as small mistakes in algebra can lead to incorrect solutions!

The quadratic equation x^2 + (k - 4)x + 2k - 11 = 0 has one double real root. Find k and solve the equation for each value of k.

Learn how to solve the quadratic equation x^2 + (k - 4)x + 2k - 11 = 0 for one double real root. Discover how to find the value of k using the discriminant condition (Δ = 0) and solve for each value of k. This solution covers algebraic concepts such as discriminants and quadratic equations, suitable for students in Grades 9-11.

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