Hence prove that the equation (x^2)-kx+2k=5 has real and distinct roots for all real values of k

We are asked to prove that the equation \(x^2 - kx + 2k = 5\) has real and distinct roots for all real values of \(k\). Let us proceed step by step to prove this.

### Step 1: Rewrite the equation in standard quadratic form
We start by rewriting the given equation \(x^2 - kx + 2k = 5\) in standard quadratic form. Subtract 5 from both sides:
\[
x^2 - kx + 2k - 5 = 0.
\]
Thus, the quadratic equation is:
\[
x^2 - kx + (2k - 5) = 0.
\]
Here, the quadratic equation is in the form \(ax^2 + bx + c = 0\), where:
- \(a = 1\),
- \(b = -k\),
- \(c = 2k - 5\).

### Step 2: Discriminant condition for real and distinct roots
For a quadratic equation to have **real and distinct roots**, the discriminant (\(\Delta\)) must be **positive**. The discriminant for a general quadratic equation \(ax^2 + bx + c = 0\) is given by:
\[
\Delta = b^2 - 4ac.
\]

Substitute the values of \(a = 1\), \(b = -k\), and \(c = 2k - 5\) into the discriminant formula:
\[
\Delta = (-k)^2 - 4(1)(2k - 5).
\]
Simplify the expression:
\[
\Delta = k^2 - 4(2k - 5).
\]
Expand the terms:
\[
\Delta = k^2 - 8k + 20.
\]

### Step 3: Analyze the discriminant
We now have the expression for the discriminant:
\[
\Delta = k^2 - 8k + 20.
\]
To ensure that the quadratic equation has real and distinct roots for all values of \(k\), we need to check if \(\Delta > 0\) for all real values of \(k\).

### Step 4: Completing the square for the discriminant
We can complete the square on the expression \(k^2 - 8k + 20\) to make the analysis easier:
\[
\Delta = (k^2 - 8k) + 20.
\]
To complete the square, take half of \(-8\), which is \(-4\), and square it to get 16. Thus:
\[
\Delta = (k - 4)^2 + 4.
\]

### Step 5: Conclusion
Since \((k - 4)^2\) is a perfect square, it is always non-negative, i.e., \((k - 4)^2 \geq 0\) for all real \(k\). Therefore, \(\Delta = (k - 4)^2 + 4 \geq 4\) for all real \(k\).

Since \(\Delta\) is always greater than 0, the quadratic equation \(x^2 - kx + 2k = 5\) has real and distinct roots for all real values of \(k\).

### Final Answer:
The equation \(x^2 - kx + 2k = 5\) has real and distinct roots for all real values of \(k\).

Would you like further details on any part of the solution, or do you have any questions?

Here are 5 questions related to this problem:
1. How do you determine the nature of roots using the discriminant?
2. What is the significance of completing the square in quadratic equations?
3. Why is \((k - 4)^2\) always non-negative?
4. Can the discriminant ever be zero in this case?
5. How do changes in the value of \(k\) affect the roots of the quadratic equation?

**Tip:** Completing the square is a useful method not only for solving quadratic equations but also for analyzing the discriminant and graphing parabolas.

Explore the proof that the quadratic equation x^2 - kx + 2k = 5 has real and distinct roots for all real values of k. This solution delves into algebraic techniques like the discriminant formula and completing the square, suitable for high school students in Grades 10-12.

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