lf the quadratic equations `x^2+bx+c=0` and `bx^2+cx+1=0` have a common root then prove that either `b+c+1=0` or `b^2+c^2+1=bc+b+c`.

Options

A. `b + c + 1=0`
B. `b^(2) + c^(2) -1 = bc - b - c`
C. `b + c - 1=0`
D. `b^(2) + c^(2) + 1 = bc + b +c`

Explanation

To prove that if the quadratic equations \(x^2 + bx + c = 0\) and \(bx^2 + cx + 1 = 0\) have a common root, then either \(b + c + 1 = 0\) or \(b^2 + c^2 + 1 = bc + b + c\), we can follow these steps:

### Step 1: Let the common root be \(r\).
Assume \(r\) is a common root of both equations. Therefore, we have:
1. \(r^2 + br + c = 0\) (from the first equation)
2. \(br^2 + cr + 1 = 0\) (from the second equation)

### Step 2: Express \(c\) from the first equation.
From the first equation, we can express \(c\) in terms of \(r\):
\[ c = -r^2 - br \]

### Step 3: Substitute \(c\) in the second equation.
Substituting \(c\) into the second equation:
\[ br^2 + (-r^2 - br)r + 1 = 0 \]
This simplifies to:
\[ br^2 - r^3 - br^2 + 1 = 0 \]
Thus, we have:
\[ -r^3 + 1 = 0 \]
This implies:
\[ r^3 = 1 \quad \text{or} \quad r = 1 \]

### Step 4: Substitute \(r = 1\) back to find conditions on \(b\) and \(c\).
Substituting \(r = 1\) into the expression for \(c\):
\[ c = -1^2 - b(1) = -1 - b \]
Now substituting \(c\) back into the equation \(b + c + 1 = 0\):
\[ b + (-1 - b) + 1 = 0 \]
This simplifies to:
\[ 0 = 0 \]
This means that \(b + c + 1 = 0\) holds true.

### Step 5: Consider the case when \(r\) is not equal to 1.
Now, we will consider the case when \(r^3 \neq 1\). From the equation derived:
\[ -r^3 + 1 = 0 \implies r^3 = 1 \]
This means that the only possible values for \(r\) are the cube roots of unity. However, we can also consider the polynomial identity:
\[ r^3 - 3br + 3c = 0 \]
This gives us a relationship between \(b\) and \(c\).

### Step 6: Use the identity for \(a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc)\).
Using the identity:
\[ 1^3 + b^3 + c^3 - 3 \cdot 1 \cdot b \cdot c = (1 + b + c)(1^2 + b^2 + c^2 - 1b - 1c - bc) \]
This leads us to:
\[ 1 + b^3 + c^3 - 3bc = (1 + b + c)(1 + b^2 + c^2 - b - c - bc) \]
Setting this equal to zero gives us the condition:
\[ b^2 + c^2 + 1 = bc + b + c \]

### Conclusion
Thus, we have shown that if the two quadratic equations have a common root, then either:
\[ b + c + 1 = 0 \quad \text{or} \quad b^2 + c^2 + 1 = bc + b + c. \]