Question

{ 2x²-y²=-1 2x+y²=17

Answer 1

2x² + 2x = 16

x² + x - 8 = 0

D = 1 + 32 = 33

x₁ = (-1 - √33)/2

x₂ = (-1 + √33)/2

y² = 17 - 2x = 17 + 1 + √33 = 18 + √33

y₁ = -√(18 + √33)

y₂ = √(18 + √33)

y² = 17 - 2x = 17 + 1 - √33 = 18 - √33

y₃ = -√(18 - √33)

y₄ = √(18 - √33)

Ответ: ( (-1 - √33)/2; -√(18 + √33)); ( (-1 - √33)/2; √(18 + √33)); ( (-1 + √33)/2; -√(18 - √33)); ( (-1 + √33)/2; √(18 - √33));

Answer 2

To solve this system of equations:

1. 2x² - y² = -1
2. 2x + y² = 17

You can use various methods, such as substitution or elimination. Here, we'll use the elimination method:

First, let's try to eliminate one of the variables. To do that, let's multiply the first equation by 2 and subtract it from the second equation:

(2x + y²) - 2(2x² - y²) = 17 - (-1)

Now, simplify both sides:

2x + y² - 4x² + 2y² = 17 + 1

Combine like terms:

2x - 4x² + 3y² = 18

Now, rewrite the equation in standard form:

-4x² + 2x + 3y² = 18

Next, we can isolate one variable from one of the equations. Let's isolate x from the first equation:

2x² - y² = -1

2x² = y² - 1

x² = (y² - 1)/2

x = ±√((y² - 1)/2)

Now, we can substitute this expression for x into the equation we just derived:

-4((y² - 1)/2) + 2(±√((y² - 1)/2)) + 3y² = 18

Now, simplify and solve for y:

-2(y² - 1) + 2(±√((y² - 1)/2)) + 3y² = 18

-2y² + 2 + 2(±√((y² - 1)/2)) + 3y² = 18

Combine like terms:

y² + 2(±√((y² - 1)/2)) = 18 - 2

y² + 2(±√((y² - 1)/2)) = 16

Now, let's solve for y. First, isolate the square root part:

2(±√((y² - 1)/2)) = 16 - y²

±√((y² - 1)/2) = (16 - y²)/2

Now, square both sides to get rid of the square root:

(y² - 1)/2 = ((16 - y²)/2)²

(y² - 1)/2 = (16 - y²)²/4

Now, cross-multiply to eliminate the fractions:

2(y² - 1) = (16 - y²)²

Expand and simplify:

2y² - 2 = (16 - y²)²

2y² - 2 = 256 - 32y² + y⁴

Now, let's rearrange and set the equation equal to 0:

y⁴ + 2y² - 32y² - 2 - 256 = 0

Combine like terms:

y⁴ - 30y² - 258 = 0

Now, you have a quartic equation in terms of y. Solving this equation may require using numerical methods, as it doesn't factor easily. You can use a graphing calculator or software like WolframAlpha to find the approximate solutions for y. Once you have those values, you can plug them back into the equation for x to find the corresponding x values.