If `c lt 0` and `ax^2 + bx + c = 0` does not have any real roots then prove that:
(i) `a-b + c lt 0`, (ii) `9a +3b +c lt 0`

Answer

To prove the statements given in the question, we start with the quadratic equation \( ax^2 + bx + c = 0 \) and the conditions \( c < 0 \) and that the equation does not have any real roots.

### Step-by-Step Solution:

1. **Understanding the Condition of No Real Roots**:
Since the quadratic equation does not have any real roots, the discriminant must be less than zero. The discriminant \( D \) for the equation \( ax^2 + bx + c = 0 \) is given by:
\[ D = b^2 - 4ac \]
Therefore, we have:
\[ b^2 - 4ac < 0 \]

2. **Finding \( f(0) \)**:
We evaluate the function at \( x = 0 \):
\[ f(0) = a(0)^2 + b(0) + c = c \]
Given that \( c < 0 \), we have:
\[ f(0) < 0 \]

3. **Evaluating \( f(-1) \)**:
Next, we evaluate the function at \( x = -1 \):
\[ f(-1) = a(-1)^2 + b(-1) + c = a - b + c \]
Since the quadratic does not intersect the x-axis and \( f(0) < 0 \), it implies that \( f(-1) < 0 \):
\[ a - b + c < 0 \]
This proves the first part:
\[ \text{(i) } a - b + c < 0 \]

4. **Evaluating \( f(3) \)**:
Now, we evaluate the function at \( x = 3 \):
\[ f(3) = a(3)^2 + b(3) + c = 9a + 3b + c \]
Again, since the quadratic does not intersect the x-axis, we have \( f(3) < 0 \):
\[ 9a + 3b + c < 0 \]
This proves the second part:
\[ \text{(ii) } 9a + 3b + c < 0 \]

### Conclusion:
Thus, we have shown that:
1. \( a - b + c < 0 \)
2. \( 9a + 3b + c < 0 \)