If the equations `ax^(2)+bx+c=0 and x^(2)+x+1=0` have a common root, then

Options

A. a+b+c=0
B. a=b=c
C. a=b or b=c or c=a
D. none of these

Answer

To solve the problem, we need to find the relationship between the coefficients \( a, b, \) and \( c \) of the quadratic equation \( ax^2 + bx + c = 0 \) and the quadratic equation \( x^2 + x + 1 = 0 \) given that they have a common root.

### Step-by-Step Solution:

1. **Identify the Roots of the Second Equation:**
The second equation is \( x^2 + x + 1 = 0 \). We can find its roots using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = 1, c = 1 \).
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2} \]
Thus, the roots are:
\[ \omega = \frac{-1 + i\sqrt{3}}{2}, \quad \omega^2 = \frac{-1 - i\sqrt{3}}{2} \]

2. **Common Root in the First Equation:**
Since the equations have a common root, we can assume that one of the roots of \( ax^2 + bx + c = 0 \) is either \( \omega \) or \( \omega^2 \).

3. **Using the Common Root:**
Let's assume \( \omega \) is a root of the first equation. Then substituting \( \omega \) into \( ax^2 + bx + c = 0 \):
\[ a\omega^2 + b\omega + c = 0 \]

4. **Substituting the Value of \( \omega^2 \):**
We know \( \omega^2 = -\omega - 1 \) (from the equation \( x^2 + x + 1 = 0 \)). Substitute this into the equation:
\[ a(-\omega - 1) + b\omega + c = 0 \]
Simplifying gives:
\[ -a\omega - a + b\omega + c = 0 \]
Rearranging terms:
\[ (b - a)\omega + (c - a) = 0 \]

5. **Setting Coefficients to Zero:**
For this equation to hold for all values of \( \omega \), both coefficients must be zero:
\[ b - a = 0 \quad \text{and} \quad c - a = 0 \]
This leads to:
\[ a = b \quad \text{and} \quad a = c \]
Therefore, we conclude:
\[ a = b = c \]

### Conclusion:
If the equations \( ax^2 + bx + c = 0 \) and \( x^2 + x + 1 = 0 \) have a common root, then it must be true that \( a = b = c \).