If the equation `ax^(2) + bx + 6 = 0` has real roots, where `a in R, b in R`, then the greatest value of 3a + b, is

Options

A. 4
B. -1
C. -2
D. 1

Explanation

To find the greatest value of \(3a + b\) given that the quadratic equation \(ax^2 + bx + 6 = 0\) has real roots, we need to follow these steps:

### Step 1: Understanding the Condition for Real Roots
The quadratic equation \(ax^2 + bx + c = 0\) has real roots if its discriminant \(D\) is greater than or equal to zero. The discriminant is given by:
\[ D = b^2 - 4ac \]
In our case, \(c = 6\). Therefore, we have:
\[ D = b^2 - 24a \geq 0 \]

### Step 2: Expressing \(3a + b\)
Let's denote \(3a + b\) as \(\lambda\). We want to maximize \(\lambda\).

### Step 3: Rearranging the Discriminant Condition
From the discriminant condition, we can rearrange it as follows:
\[ b^2 \geq 24a \]
Now, substituting \(b\) in terms of \(\lambda\):
\[ b = \lambda - 3a \]
Substituting this into the discriminant condition gives:
\[ (\lambda - 3a)^2 \geq 24a \]

### Step 4: Expanding and Rearranging
Expanding the left side:
\[ \lambda^2 - 6a\lambda + 9a^2 \geq 24a \]
Rearranging gives:
\[ 9a^2 - 6a\lambda + (\lambda^2 - 24a) \geq 0 \]
This simplifies to:
\[ 9a^2 - (6\lambda + 24)a + \lambda^2 \geq 0 \]

### Step 5: Finding the Discriminant of the New Quadratic
For this quadratic in \(a\) to have real solutions, its discriminant must be non-negative:
\[ D_a = (6\lambda + 24)^2 - 4 \cdot 9 \cdot \lambda^2 \geq 0 \]
Calculating the discriminant:
\[ D_a = 36\lambda^2 + 288\lambda + 576 - 36\lambda^2 \geq 0 \]
This simplifies to:
\[ 288\lambda + 576 \geq 0 \]

### Step 6: Solving for \(\lambda\)
Dividing the entire inequality by 288:
\[ \lambda + 2 \geq 0 \]
Thus, we find:
\[ \lambda \geq -2 \]

### Step 7: Conclusion
The greatest value of \(3a + b\) is \(-2\).

### Final Answer
The greatest value of \(3a + b\) is \(\boxed{-2}\).

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