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!