The quadratic equation `((x+b)(x+c))/((b-a)(c-a))+((x+c)(x+a))/((c-b)(a-b))+((x+a)(x+b))/((a-c)(b-c))=1` has (A) Two real and distinct roots (B) Two Equal roots (C) Non Real Complex Roots (D) Infinite roots

Options

A. Two real and distinct roots
B. Two equal roots
C. Non-real complex roots
D. Infinite roots

Answer

To solve the given quadratic equation

\[ \frac{(x+b)(x+c)}{(b-a)(c-a)} + \frac{(x+c)(x+a)}{(c-b)(a-b)} + \frac{(x+a)(x+b)}{(a-c)(b-c)} = 1, \]

we will analyze the equation step by step.

### Step 1: Substitute Values
Let's substitute \( x = -a \) into the equation.

\[ \frac{(-a+b)(-a+c)}{(b-a)(c-a)} + \frac{(-a+c)(-a+a)}{(c-b)(a-b)} + \frac{(-a+a)(-a+b)}{(a-c)(b-c)} = 1. \]

The second term becomes zero because \((-a + a) = 0\). The third term also becomes zero for the same reason. Thus, we only need to evaluate the first term:

\[ \frac{(-a+b)(-a+c)}{(b-a)(c-a)} = 1. \]

### Step 2: Simplify the First Term
This simplifies to:

\[ \frac{(b-a)(c-a)}{(b-a)(c-a)} = 1. \]

This shows that the equation holds true for \( x = -a \).

### Step 3: Substitute Another Value
Now, let's substitute \( x = -b \):

\[ \frac{(-b+b)(-b+c)}{(b-a)(c-a)} + \frac{(-b+c)(-b+a)}{(c-b)(a-b)} + \frac{(-b+a)(-b+b)}{(a-c)(b-c)} = 1. \]

Again, the first term becomes zero, and the third term also becomes zero. Therefore, we only need to evaluate the second term:

\[ \frac{(-b+c)(-b+a)}{(c-b)(a-b)} = 1. \]

### Step 4: Simplify the Second Term
This simplifies to:

\[ \frac{(c-b)(a-b)}{(c-b)(a-b)} = 1. \]

This shows that the equation holds true for \( x = -b \).

### Step 5: Substitute the Last Value
Next, substitute \( x = -c \):

\[ \frac{(-c+b)(-c+c)}{(b-a)(c-a)} + \frac{(-c+c)(-c+a)}{(c-b)(a-b)} + \frac{(-c+a)(-c+b)}{(a-c)(b-c)} = 1. \]

Again, the first term becomes zero, and the second term also becomes zero. Therefore, we only need to evaluate the third term:

\[ \frac{(-c+a)(-c+b)}{(a-c)(b-c)} = 1. \]

### Step 6: Simplify the Third Term
This simplifies to:

\[ \frac{(a-c)(b-c)}{(a-c)(b-c)} = 1. \]

This shows that the equation holds true for \( x = -c \).

### Conclusion
We have found three values \( x = -a, -b, -c \) that satisfy the equation. However, since this is a quadratic equation, it can have at most two roots. The fact that we have found three distinct values indicates that the equation is an identity, meaning it holds true for all \( x \).

Thus, the equation has infinite roots.

### Final Answer
The correct option is (D) Infinite roots.

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