If `ax^(2)-bx+5=0` does not have two distinct real roots, then find the minimun value of 5a+b.

Answer

To solve the problem, we need to determine the minimum value of \( 5a + b \) given that the quadratic equation \( ax^2 - bx + 5 = 0 \) does not have two distinct real roots. This means that the discriminant of the quadratic must be less than or equal to zero.

### Step-by-Step Solution:

1. **Identify the Discriminant**:
The discriminant \( D \) of the quadratic equation \( ax^2 - bx + 5 = 0 \) is given by:
\[ D = (-b)^2 - 4 \cdot a \cdot 5 = b^2 - 20a \]
For the equation to not have two distinct real roots, we require:
\[ D \leq 0 \implies b^2 - 20a \leq 0 \]
This simplifies to:
\[ b^2 \leq 20a \]

2. **Express \( 5a + b \)**:
We want to find the minimum value of \( 5a + b \). We can express \( 5a + b \) in terms of \( b \):
\[ 5a + b = 5a + b \]

3. **Substitute for \( a \)**:
From the inequality \( b^2 \leq 20a \), we can express \( a \) in terms of \( b \):
\[ a \geq \frac{b^2}{20} \]
Substituting this into \( 5a + b \):
\[ 5a + b \geq 5\left(\frac{b^2}{20}\right) + b = \frac{b^2}{4} + b \]

4. **Minimize the Expression**:
Let \( z = \frac{b^2}{4} + b \). To find the minimum value of \( z \), we can take the derivative and set it to zero:
\[ \frac{dz}{db} = \frac{2b}{4} + 1 = \frac{b}{2} + 1 \]
Setting the derivative to zero for minimization:
\[ \frac{b}{2} + 1 = 0 \implies b = -2 \]

5. **Find Corresponding \( a \)**:
Substitute \( b = -2 \) back into the inequality to find \( a \):
\[ (-2)^2 \leq 20a \implies 4 \leq 20a \implies a \geq \frac{4}{20} = \frac{1}{5} \]

6. **Calculate Minimum Value of \( 5a + b \)**:
Substitute \( b = -2 \) and \( a = \frac{1}{5} \) into \( 5a + b \):
\[ 5a + b = 5\left(\frac{1}{5}\right) - 2 = 1 - 2 = -1 \]

Thus, the minimum value of \( 5a + b \) is \(-1\).

### Final Answer:
The minimum value of \( 5a + b \) is \(-1\).