If `ax^(2) + 2bx + c = 0 and ax^(2) + 2cx + b = 0, b ne c` have a common root, then `(a + b + c)/( a)` is equal to

Options

A. `-2`
B. `-1`
C. `3//4`
D. `-1//4`

Answer

To solve the problem step by step, we need to find the value of \((a + b + c)/a\) given that the equations \(ax^2 + 2bx + c = 0\) and \(ax^2 + 2cx + b = 0\) have a common root.

### Step 1: Set up the equations
We have two quadratic equations:
1. \(ax^2 + 2bx + c = 0\) (Equation 1)
2. \(ax^2 + 2cx + b = 0\) (Equation 2)

### Step 2: Subtract the two equations
Subtract Equation 2 from Equation 1:
\[ (ax^2 + 2bx + c) - (ax^2 + 2cx + b) = 0 \]
This simplifies to:
\[ 2bx - 2cx + c - b = 0 \]
Rearranging gives:
\[ 2(b - c)x + (c - b) = 0 \]

### Step 3: Factor the equation
Factoring out common terms:
\[ 2(b - c)x + (c - b) = 0 \]
This can be rewritten as:
\[ (b - c)(2x - 1) = 0 \]

### Step 4: Solve for \(x\)
Since \(b \neq c\), we can conclude:
\[ 2x - 1 = 0 \implies x = \frac{1}{2} \]

### Step 5: Substitute \(x\) back into one of the equations
Substituting \(x = \frac{1}{2}\) into Equation 1:
\[ a\left(\frac{1}{2}\right)^2 + 2b\left(\frac{1}{2}\right) + c = 0 \]
This simplifies to:
\[ a\left(\frac{1}{4}\right) + b + c = 0 \]
Multiplying through by 4 to eliminate the fraction:
\[ a + 4b + 4c = 0 \]
Thus, we can express \(b + c\) in terms of \(a\):
\[ 4b + 4c = -a \implies b + c = -\frac{a}{4} \]

### Step 6: Substitute \(b + c\) into \((a + b + c)/a\)
Now we need to find \(\frac{a + b + c}{a}\):
\[ a + b + c = a + \left(-\frac{a}{4}\right) = a - \frac{a}{4} = \frac{4a - a}{4} = \frac{3a}{4} \]
So,
\[ \frac{a + b + c}{a} = \frac{\frac{3a}{4}}{a} = \frac{3}{4} \]

### Final Answer
The value of \(\frac{a + b + c}{a}\) is \(\frac{3}{4}\).