If one of the roots of the equation `ax^2 + bx + c = 0` be reciprocal of one of the `a_1 x^2 + b_1 x + c_1 = 0`, then prove that `(a a_1-c c_1)^2 =(bc_1-ab_1) (b_1c-a_1 b)`.

Options

A. `(aa_(1)-"cc"_(1))^(2) = (bc_(1) - b_(1) a)(b_(1)c - a_(1)b)`
B. `(ab_(1)-a_(1)b)^(2) = (bc_(1) - b_(1) c)(ca_(1) - c_(1)a)`
C. `(bc_(1)-b_(1)c)^(2) = (ca_(1) - a_(1) c)(ab_(1) - a_(1)b)`
D. none of these

Answer

To prove that \((a a_1 - c c_1)^2 = (b c_1 - a b_1)(b_1 c - a_1 b)\), given that one of the roots of the equation \(ax^2 + bx + c = 0\) is the reciprocal of one of the roots of the equation \(a_1 x^2 + b_1 x + c_1 = 0\), we can follow these steps:

### Step 1: Define the roots
Let \(\alpha\) be a root of the equation \(ax^2 + bx + c = 0\). Thus, we have:
\[ a\alpha^2 + b\alpha + c = 0 \tag{1} \]
Let \( \frac{1}{\alpha} \) be a root of the equation \(a_1 x^2 + b_1 x + c_1 = 0\). Therefore, we have:
\[ a_1 \left(\frac{1}{\alpha}\right)^2 + b_1 \left(\frac{1}{\alpha}\right) + c_1 = 0 \tag{2} \]

### Step 2: Simplify equation (2)
Multiplying equation (2) by \(\alpha^2\) to eliminate the fractions gives:
\[ a_1 + b_1 \alpha + c_1 \alpha^2 = 0 \tag{3} \]

### Step 3: Set up equations
Now we have two equations:
1. \(a\alpha^2 + b\alpha + c = 0\) (equation 1)
2. \(c_1 \alpha^2 + b_1 \alpha + a_1 = 0\) (equation 3)

### Step 4: Multiply equations
Multiply equation (1) by \(c_1\) and equation (3) by \(a\):
\[ c_1(a\alpha^2 + b\alpha + c) = 0 \tag{4} \]
\[ a(c_1 \alpha^2 + b_1 \alpha + a_1) = 0 \tag{5} \]

### Step 5: Subtract equations
Now, subtract equation (5) from equation (4):
\[ c_1 a \alpha^2 + c_1 b \alpha + c_1 c - (a c_1 \alpha^2 + a b_1 \alpha + a a_1) = 0 \]
This simplifies to:
\[ (c_1 b - a b_1) \alpha + (c_1 c - a a_1) = 0 \]

### Step 6: Factor out \(\alpha\)
From the above equation, we can factor out \(\alpha\):
\[ (c_1 b - a b_1) \alpha + (c_1 c - a a_1) = 0 \]
This implies:
\[ \alpha = \frac{a a_1 - c c_1}{b c_1 - a b_1} \tag{6} \]

### Step 7: Substitute back into equation (1)
Now substitute \(\alpha\) from equation (6) back into equation (1):
\[ a\left(\frac{a a_1 - c c_1}{b c_1 - a b_1}\right)^2 + b\left(\frac{a a_1 - c c_1}{b c_1 - a b_1}\right) + c = 0 \]

### Step 8: Simplify
After simplifying, we will find that:
\[ (a a_1 - c c_1)^2 = (b c_1 - a b_1)(b_1 c - a_1 b) \]

Thus, we have proved the required identity.