If roots of the equation `(a^(2)+b^(2))x^(2)-2(ac+bd)x+(c^(2)+d^(2))=0` are equal, then `bc-ad` = ______

Options

A. 1
B. 0
C. `-1`
D. 2

Answer

To solve the problem, we need to determine the condition under which the roots of the given quadratic equation are equal. The equation is:

\[ (a^2 + b^2)x^2 - 2(ac + bd)x + (c^2 + d^2) = 0 \]

### Step 1: Identify the coefficients
In a standard quadratic equation of the form \(Ax^2 + Bx + C = 0\), we can identify:
- \(A = a^2 + b^2\)
- \(B = -2(ac + bd)\)
- \(C = c^2 + d^2\)

### Step 2: Set the discriminant to zero
For the roots of the quadratic equation to be equal, the discriminant must be zero. The discriminant \(D\) is given by:

\[ D = B^2 - 4AC \]

Substituting in our coefficients:

\[ D = (-2(ac + bd))^2 - 4(a^2 + b^2)(c^2 + d^2) \]

### Step 3: Simplify the discriminant
Calculating \(D\):

\[ D = 4(ac + bd)^2 - 4(a^2 + b^2)(c^2 + d^2) \]

Factoring out the 4:

\[ D = 4\left((ac + bd)^2 - (a^2 + b^2)(c^2 + d^2)\right) \]

### Step 4: Set the discriminant equal to zero
For the roots to be equal, we set \(D = 0\):

\[ (ac + bd)^2 - (a^2 + b^2)(c^2 + d^2) = 0 \]

### Step 5: Rearranging the equation
This can be rearranged to:

\[ (ac + bd)^2 = (a^2 + b^2)(c^2 + d^2) \]

### Step 6: Use the identity
Using the identity \(x^2 + y^2 = (x+y)^2 - 2xy\), we can express:

\[ (ac + bd)^2 = a^2c^2 + b^2d^2 + 2abcd \]

Thus, we have:

\[ a^2c^2 + b^2d^2 + 2abcd = a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2 \]

### Step 7: Simplifying further
By simplifying, we can rearrange terms:

\[ 0 = a^2d^2 + b^2c^2 - 2abcd \]

### Step 8: Factor the equation
This can be factored as:

\[ 0 = (bc - ad)^2 \]

### Step 9: Conclusion
Since the square of a real number is zero only when the number itself is zero, we conclude:

\[ bc - ad = 0 \]

Thus, the answer is:

\[ \boxed{0} \]