If `a + b + c >(9c)/4` and quadratic equation `ax^2 + 2bx-5c = 0` has non-real roots, then-

Options

A. `a gt 0 , c gt 0`
B. `a gt 0 , c lt 0`
C. `a lt 0 , c lt 0 `
D. ` a lt 0 , c gt 0`

Answer

To solve the problem step by step, we need to analyze the given conditions and derive the necessary inequalities.

### Step 1: Analyze the first inequality
We are given the inequality:
\[ a + b + c > \frac{9c}{4} \]

To simplify this, we can rearrange the terms:
\[ a + b + c - \frac{9c}{4} > 0 \]

### Step 2: Combine like terms
To combine the terms involving \(c\), we can express \(c\) in terms of a common denominator:
\[ a + b + c - \frac{9c}{4} = a + b + \frac{4c}{4} - \frac{9c}{4} = a + b - \frac{5c}{4} > 0 \]

Multiplying through by 4 to eliminate the fraction gives:
\[ 4a + 4b - 5c > 0 \]

### Step 3: Analyze the quadratic equation
The quadratic equation given is:
\[ ax^2 + 2bx - 5c = 0 \]

For this quadratic to have non-real roots, the discriminant must be less than zero. The discriminant \(D\) for a quadratic equation \(Ax^2 + Bx + C = 0\) is given by:
\[ D = B^2 - 4AC \]

In our case:
- \(A = a\)
- \(B = 2b\)
- \(C = -5c\)

Thus, the discriminant becomes:
\[ D = (2b)^2 - 4(a)(-5c) = 4b^2 + 20ac \]

### Step 4: Set up the inequality for non-real roots
For the roots to be non-real, we require:
\[ 4b^2 + 20ac < 0 \]

### Step 5: Rearranging the inequality
We can factor out 4 from the left-hand side:
\[ 4(b^2 + 5ac) < 0 \]

Dividing both sides by 4 (which does not change the inequality since 4 is positive):
\[ b^2 + 5ac < 0 \]

### Step 6: Summary of conditions
We now have two inequalities to consider:
1. \( 4a + 4b - 5c > 0 \)
2. \( b^2 + 5ac < 0 \)

### Conclusion
From the inequalities derived, we conclude that:
- \( a + b + c > \frac{9c}{4} \) implies \( 4a + 4b > 5c \)
- The quadratic equation \( ax^2 + 2bx - 5c = 0 \) has non-real roots if \( b^2 + 5ac < 0 \)