The equation x^2 - 6x + k = 0 has roots that are both positive. What is the range of k?

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Q: The equation x^2 - 6x + k = 0 has roots that are both positive. What is the range of k?

Solution: For both roots to be positive, k must be greater than the square of half the coefficient of x: k > (6/2)^2 = 9.

Steps: 13

Step 1: Identify the equation given, which is x^2 - 6x + k = 0.
Step 2: Understand that the roots of a quadratic equation are the values of x that make the equation equal to zero.
Step 3: Recall that for a quadratic equation ax^2 + bx + c = 0, the roots can be found using the quadratic formula: x = (-b ± √(b² - 4ac)) / (2a).
Step 4: In our equation, a = 1, b = -6, and c = k.
Step 5: For the roots to be positive, we need to ensure two conditions: the sum of the roots must be positive, and the product of the roots must be positive.
Step 6: The sum of the roots is given by -b/a, which is 6/1 = 6. This is already positive.
Step 7: The product of the roots is given by c/a, which is k/1 = k. For the product to be positive, k must be greater than 0.
Step 8: Additionally, we need to ensure that the roots are not only positive but also real. This requires the discriminant (b² - 4ac) to be non-negative.
Step 9: Calculate the discriminant: (-6)² - 4(1)(k) = 36 - 4k.
Step 10: Set the discriminant greater than or equal to zero: 36 - 4k ≥ 0.
Step 11: Solve for k: 36 ≥ 4k, which simplifies to k ≤ 9.
Step 12: Combine the conditions: k must be greater than 0 (for positive product) and less than or equal to 9 (for real roots).
Step 13: Therefore, the range of k is 0 < k ≤ 9.