If `x^(2)+y^(2)=25, xy=12` then x=

Options

A. {3,4}
B. {3,-3}
C. {3,4,-3,-4}
D. {-3,-3}

Answer

To solve the problem given \( x^2 + y^2 = 25 \) and \( xy = 12 \), we can follow these steps:

### Step 1: Use the identity for the square of a sum
We know that:
\[ (x + y)^2 = x^2 + y^2 + 2xy \]
Substituting the known values:
\[ (x + y)^2 = 25 + 2(12) = 25 + 24 = 49 \]

### Step 2: Solve for \( x + y \)
Taking the square root of both sides:
\[ x + y = \pm 7 \]

### Step 3: Use the identity for the square of a difference
We also know that:
\[ (x - y)^2 = x^2 + y^2 - 2xy \]
Substituting the known values:
\[ (x - y)^2 = 25 - 2(12) = 25 - 24 = 1 \]

### Step 4: Solve for \( x - y \)
Taking the square root of both sides:
\[ x - y = \pm 1 \]

### Step 5: Set up the equations
Now we have two sets of equations:
1. \( x + y = 7 \) and \( x - y = 1 \)
2. \( x + y = 7 \) and \( x - y = -1 \)
3. \( x + y = -7 \) and \( x - y = 1 \)
4. \( x + y = -7 \) and \( x - y = -1 \)

### Step 6: Solve the first set of equations
Using the first set:
1. \( x + y = 7 \)
2. \( x - y = 1 \)

Adding these two equations:
\[ (x + y) + (x - y) = 7 + 1 \implies 2x = 8 \implies x = 4 \]

Now substitute \( x = 4 \) back into \( x + y = 7 \):
\[ 4 + y = 7 \implies y = 3 \]

### Step 7: Solve the second set of equations
Using the second set:
1. \( x + y = 7 \)
2. \( x - y = -1 \)

Adding these two equations:
\[ (x + y) + (x - y) = 7 - 1 \implies 2x = 6 \implies x = 3 \]

Now substitute \( x = 3 \) back into \( x + y = 7 \):
\[ 3 + y = 7 \implies y = 4 \]

### Step 8: Solve the third set of equations
Using the third set:
1. \( x + y = -7 \)
2. \( x - y = 1 \)

Adding these two equations:
\[ (x + y) + (x - y) = -7 + 1 \implies 2x = -6 \implies x = -3 \]

Now substitute \( x = -3 \) back into \( x + y = -7 \):
\[ -3 + y = -7 \implies y = -4 \]

### Step 9: Solve the fourth set of equations
Using the fourth set:
1. \( x + y = -7 \)
2. \( x - y = -1 \)

Adding these two equations:
\[ (x + y) + (x - y) = -7 - 1 \implies 2x = -8 \implies x = -4 \]

Now substitute \( x = -4 \) back into \( x + y = -7 \):
\[ -4 + y = -7 \implies y = -3 \]

### Final Values of \( x \)
The possible values of \( x \) are \( 4, 3, -3, -4 \).

### Conclusion
The values of \( x \) are \( 4 \) and \( 3 \) (the positive values).