Question

[x² + y² = 25, xy=-12​

Answer 1

Объяснение:

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Answer 2

To solve the system of equations x² + y² = 25 and xy = -12, we can use the method of substitution.

From the second equation, we can express one variable in terms of the other. Solving for x or y, we have:

xy = -12 x = -12/y

Substituting this expression into the first equation, we get:

(-12/y)² + y² = 25 144/y² + y² = 25

Multiplying through by y² to eliminate the denominator, we have:

144 + y⁴ = 25y²

Rearranging the equation:

y⁴ - 25y² + 144 = 0

This is a quadratic equation in terms of y². We can substitute u = y² to simplify it:

u² - 25u + 144 = 0

Now, we can solve this quadratic equation for u. Factoring it, we have:

(u - 9)(u - 16) = 0

Setting each factor equal to zero, we get:

u - 9 = 0 --> u = 9 u - 16 = 0 --> u = 16

Substituting u back as y², we have:

y² = 9 --> y = ±√9 = ±3 y² = 16 --> y = ±√16 = ±4

Now, substituting the values of y back into the equation xy = -12, we can solve for x:

For y = 3: x(3) = -12 --> x = -12/3 = -4

For y = -3: x(-3) = -12 --> x = -12/(-3) = 4

For y = 4: x(4) = -12 --> x = -12/4 = -3

For y = -4: x(-4) = -12 --> x = -12/(-4) = 3

Therefore, the solution to the system of equations is: (x, y) = {(-4, 3), (4, -3), (-3, 4), (3, -4)}