Which one of the following is one of the roots of the equation `(b-c)^(x^(2)) + (c-a)x + ( a-b) = 0` ?

Options

A. `(c-a)//(b-c)`
B. `(a-b)//(b-c)`
C. `(b-c)//(a-b)`
D. `(c-a)//(a-b)`

Explanation

To solve the equation \((b-c)x^2 + (c-a)x + (a-b) = 0\) and find one of its roots, we can follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation is of the form \(Ax^2 + Bx + C = 0\), where:
- \(A = b - c\)
- \(B = c - a\)
- \(C = a - b\)

### Step 2: Apply the quadratic formula
The roots of a quadratic equation can be found using the quadratic formula:
\[ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]
Substituting the values of \(A\), \(B\), and \(C\):
\[ x = \frac{-(c-a) \pm \sqrt{(c-a)^2 - 4(b-c)(a-b)}}{2(b-c)} \]

### Step 3: Simplify the expression
First, calculate the discriminant:
\[ D = (c-a)^2 - 4(b-c)(a-b) \]
Now, substituting \(D\) back into the quadratic formula:
\[ x = \frac{a-c \pm \sqrt{(c-a)^2 - 4(b-c)(a-b)}}{2(b-c)} \]

### Step 4: Factor the equation
To find the roots more easily, we can factor the quadratic expression. We look for two numbers that multiply to \(A \cdot C = (b-c)(a-b)\) and add to \(B = (c-a)\).

### Step 5: Set up the factors
We can express the quadratic as:
\[ (b-c)x^2 + (c-a)x + (a-b) = 0 \]
This can be factored into:
\[ (b-c)(x - r_1)(x - r_2) = 0 \]
where \(r_1\) and \(r_2\) are the roots.

### Step 6: Solve for the roots
From the factorization, we can find the roots:
1. One root is \(x = 1\).
2. The other root can be derived from the equation \(b - c = a - b\).

### Conclusion
Thus, one of the roots of the equation is:
\[ x = 1 \]
And the other root can be expressed as:
\[ x = \frac{a - b}{b - c} \]