Given a,b, c are three distinct real numbers satisfying the inequality `a-2b+4c gt0` and the equation `ax ^(2) + bx +c =0` has no real roots. Then the possible value (s) of `(4a +2b+c)/(a+3b+9c)` is/are:

Options

A. 2
B. `-1`
C. 3
D. `sqrt2`

Answer

To solve the problem step-by-step, we need to analyze the given conditions and derive the possible values of the expression \((4a + 2b + c)/(a + 3b + 9c)\).

### Step 1: Understand the conditions
We have two conditions:
1. The inequality \(a - 2b + 4c > 0\).
2. The quadratic equation \(ax^2 + bx + c = 0\) has no real roots, which implies that its discriminant \(D < 0\).

### Step 2: Write the discriminant condition
The discriminant \(D\) for the quadratic equation \(ax^2 + bx + c = 0\) is given by:
\[ D = b^2 - 4ac \]
Since the equation has no real roots, we have:
\[ b^2 - 4ac < 0 \quad \text{(1)} \]

### Step 3: Analyze the inequality
The inequality \(a - 2b + 4c > 0\) can be rearranged to express \(a\) in terms of \(b\) and \(c\):
\[ a > 2b - 4c \quad \text{(2)} \]

### Step 4: Substitute values into the expression
We need to evaluate the expression:
\[ \frac{4a + 2b + c}{a + 3b + 9c} \]

### Step 5: Let’s denote the expression as \(k\)
Let:
\[ k = \frac{4a + 2b + c}{a + 3b + 9c} \]
We can cross-multiply to eliminate the fraction:
\[ k(a + 3b + 9c) = 4a + 2b + c \]
This simplifies to:
\[ ka + 3kb + 9kc = 4a + 2b + c \]
Rearranging gives:
\[ (ka - 4a) + (3kb - 2b) + (9kc - c) = 0 \]
Factoring out the terms:
\[ a(k - 4) + b(3k - 2) + c(9k - 1) = 0 \quad \text{(3)} \]

### Step 6: Analyze the coefficients
For the equation (3) to hold true for distinct \(a\), \(b\), and \(c\), the coefficients must be such that:
\[ k - 4 = 0 \quad \text{or} \quad 3k - 2 = 0 \quad \text{or} \quad 9k - 1 = 0 \]

### Step 7: Solve for \(k\)
1. From \(k - 4 = 0\):
\[ k = 4 \]
2. From \(3k - 2 = 0\):
\[ 3k = 2 \implies k = \frac{2}{3} \]
3. From \(9k - 1 = 0\):
\[ 9k = 1 \implies k = \frac{1}{9} \]

### Step 8: Conclusion
The possible values of \(k\) are:
\[ k = 4, \quad k = \frac{2}{3}, \quad k = \frac{1}{9} \]

### Final Answer
Thus, the possible values of \(\frac{4a + 2b + c}{a + 3b + 9c}\) are \(4\), \(\frac{2}{3}\), and \(\frac{1}{9}\).