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 conditions under which the quadratic equation \( ax^2 - bx + 5 = 0 \) does not have two distinct real roots. This occurs when the discriminant of the quadratic equation is less than or equal to zero.

### Step 1: Write the discriminant condition
The discriminant \( D \) of the quadratic equation \( ax^2 - bx + 5 = 0 \) is given by:
\[ D = b^2 - 4ac \]
Here, \( a = a \), \( b = -b \), and \( c = 5 \). Therefore, the discriminant becomes:
\[ D = (-b)^2 - 4a \cdot 5 = b^2 - 20a \]
For the equation to not have two distinct real roots, we need:
\[ D \leq 0 \implies b^2 - 20a \leq 0 \]
This simplifies to:
\[ b^2 \leq 20a \]

### Step 2: Rearranging the inequality
We can rearrange the inequality to express \( a \) in terms of \( b \):
\[ a \geq \frac{b^2}{20} \]

### Step 3: Find the expression for \( 5a + b \)
We want to minimize the expression \( 5a + b \). Substituting \( a \) from the inequality gives:
\[ 5a + b \geq 5\left(\frac{b^2}{20}\right) + b = \frac{b^2}{4} + b \]

### Step 4: Minimize the function \( f(b) = \frac{b^2}{4} + b \)
To find the minimum value of \( f(b) \), we can take the derivative and set it to zero:
\[ f'(b) = \frac{1}{2}b + 1 \]
Setting \( f'(b) = 0 \):
\[ \frac{1}{2}b + 1 = 0 \implies b = -2 \]

### Step 5: Find the corresponding value of \( a \)
Substituting \( b = -2 \) back into the inequality for \( a \):
\[ a \geq \frac{(-2)^2}{20} = \frac{4}{20} = \frac{1}{5} \]

### Step 6: Calculate \( 5a + b \)
Now substituting \( a = \frac{1}{5} \) and \( b = -2 \) into \( 5a + b \):
\[ 5a + b = 5\left(\frac{1}{5}\right) + (-2) = 1 - 2 = -1 \]

### Conclusion
Thus, the minimum value of \( 5a + b \) is:
\[ \boxed{-1} \]