Find the range of values of `'a'` so that the equation `x^2-(a-5)x+(a-15/4)=0` has at least one positive root

Answer

To find the range of values of \( a \) such that the equation

\[ x^2 - (a - 5)x + \left(a - \frac{15}{4}\right) = 0 \]

has at least one positive root, we will analyze the conditions under which this quadratic equation has positive roots.

### Step 1: Identify the coefficients
The quadratic equation can be rewritten in standard form as:

\[ Ax^2 + Bx + C = 0 \]

where:
- \( A = 1 \)
- \( B = -(a - 5) = 5 - a \)
- \( C = a - \frac{15}{4} \)

### Step 2: Condition for real roots
For the quadratic equation to have real roots, the discriminant \( D \) must be greater than or equal to zero:

\[ D = B^2 - 4AC \]

Calculating the discriminant:

\[ D = (5 - a)^2 - 4 \cdot 1 \cdot \left(a - \frac{15}{4}\right) \]

Expanding this:

\[ D = (5 - a)^2 - 4a + 15 \]
\[ D = 25 - 10a + a^2 - 4a + 15 \]
\[ D = a^2 - 14a + 40 \]

Setting the discriminant greater than zero for at least one positive root:

\[ a^2 - 14a + 40 > 0 \]

### Step 3: Solve the inequality
To solve the quadratic inequality \( a^2 - 14a + 40 > 0 \), we first find the roots of the equation \( a^2 - 14a + 40 = 0 \) using the quadratic formula:

\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{14 \pm \sqrt{(-14)^2 - 4 \cdot 1 \cdot 40}}{2 \cdot 1} \]
\[ = \frac{14 \pm \sqrt{196 - 160}}{2} = \frac{14 \pm \sqrt{36}}{2} = \frac{14 \pm 6}{2} \]

Calculating the roots:

\[ a_1 = \frac{20}{2} = 10, \quad a_2 = \frac{8}{2} = 4 \]

### Step 4: Analyze the intervals
The roots divide the number line into intervals. We test each interval to determine where the inequality holds:

1. \( (-\infty, 4) \)
2. \( (4, 10) \)
3. \( (10, \infty) \)

Testing the intervals:
- For \( a < 4 \) (e.g., \( a = 0 \)): \( 0^2 - 14(0) + 40 = 40 > 0 \) (True)
- For \( 4 < a < 10 \) (e.g., \( a = 5 \)): \( 5^2 - 14(5) + 40 = 25 - 70 + 40 = -5 < 0 \) (False)
- For \( a > 10 \) (e.g., \( a = 11 \)): \( 11^2 - 14(11) + 40 = 121 - 154 + 40 = 7 > 0 \) (True)

Thus, the solution to the inequality \( a^2 - 14a + 40 > 0 \) is:

\[ a \in (-\infty, 4) \cup (10, \infty) \]

### Step 5: Condition for positive roots
For the quadratic to have at least one positive root, we also need to consider the product of the roots, which is given by:

\[ \text{Product of roots} = \frac{C}{A} = a - \frac{15}{4} \]

For at least one root to be positive, we require:

\[ a - \frac{15}{4} < 0 \implies a < \frac{15}{4} \]

### Step 6: Combine conditions
Now we combine the conditions:

1. From the discriminant: \( a \in (-\infty, 4) \cup (10, \infty) \)
2. From the product of roots: \( a < \frac{15}{4} \)

The intersection of these conditions gives:

\[ a \in (-\infty, 4) \cup (10, \infty) \cap (-\infty, \frac{15}{4}) \]

Since \( \frac{15}{4} = 3.75 \), the final range of \( a \) that satisfies both conditions is:

\[ a \in (-\infty, 4) \text{ (since } 4 > \frac{15}{4} \text{)} \]

Thus, the final answer is:

\[ \boxed{(-\infty, 4)} \]