Let `a ,b ,and c` be real numbers such that `4a+2b+c=0 and a b > 0.` Then the equation `a x^2+b x+c = 0

Options

A. real roots
B. complex roots
C. exactly one root
D. none of these

Answer

To solve the problem, we need to analyze the given conditions and the quadratic equation.

### Step 1: Understand the given equations
We have the quadratic equation:
\[ ax^2 + bx + c = 0 \]
And we are given the condition:
\[ 4a + 2b + c = 0 \]
along with the condition that \( ab > 0 \).

### Step 2: Substitute \( x = 2 \) into the quadratic equation
Let's substitute \( x = 2 \) into the quadratic equation:
\[ a(2^2) + b(2) + c = 0 \]
This simplifies to:
\[ 4a + 2b + c = 0 \]

### Step 3: Relate the two equations
From the substitution, we see that:
\[ 4a + 2b + c = 0 \]
This is exactly the same as the given condition \( 4a + 2b + c = 0 \). Therefore, \( x = 2 \) is a root of the quadratic equation.

### Step 4: Analyze the condition \( ab > 0 \)
The condition \( ab > 0 \) implies that both \( a \) and \( b \) are either positive or negative. This means that the parabola represented by the quadratic equation opens upwards (if \( a > 0 \)) or downwards (if \( a < 0 \)).

### Step 5: Determine the nature of the roots
Since we have found one root \( x = 2 \), and because the coefficients \( a \) and \( b \) are such that \( ab > 0 \), the quadratic must have another real root. This is because the roots of a quadratic equation either both are real or both are imaginary, and since we have one real root, the other must also be real.

### Conclusion
Thus, we conclude that both roots of the quadratic equation \( ax^2 + bx + c = 0 \) are real.

### Final Answer
Both roots of the equation are real.

---