Question

Найдите все отрицательные значения параметра а, при которых система уравнения {█(2√(y^2-4y+4)+3|x|=11-y,@〖25x〗^2-20ax=y^2-〖4a〗^2 )┤ имеет одно решение. Если такое значение одно, то запишите его в ответ. Если таких значений несколько, то в ответ записать его сумму.

Answer 1

Всё очень просто. Правда непонятно зачем дано три системы, если всего неизвестных две.
Но наверное решается так, сначала решается эта пара уравнений:
3x-2y=7
x+y=4

3x-2y=7
2x+2y=8
Тогда х=3, у=1
Теперь эти значения подставляем в третье уравнение:
2x-y=p
2*3-1=р
р=5
Ответ: при р=5 система имеет решение.

Answer 2

Finding Negative Values of Parameter a for a System of Equations

To find the negative values of parameter a for which the system of equations has exactly one solution, let's analyze the given system of equations:

Equation 1: 2√(y^2-4y+4) + 3|x| = 11 - y Equation 2: 25x^2 - 20ax = y^2 - 4a^2

We want to find the values of parameter a that result in a unique solution for this system.

To begin, let's consider Equation 1. The expression 2√(y^2-4y+4) involves the square root of a quadratic expression. For a real solution, the discriminant of the quadratic expression must be greater than or equal to zero.

The discriminant of the quadratic expression y^2-4y+4 is given by b^2 - 4ac, where a = 1, b = -4, and c = 4. Evaluating the discriminant, we have:

(-4)^2 - 4(1)(4) = 16 - 16 = 0

Since the discriminant is equal to zero, the quadratic expression y^2-4y+4 has a double root. This means that Equation 1 has a unique solution only when the expression inside the square root is equal to zero.

Now, let's consider Equation 2. We want to find the values of parameter a that result in a unique solution for this equation.

The equation 25x^2 - 20ax = y^2 - 4a^2 represents a quadratic equation in terms of x. For a unique solution, the discriminant of this quadratic equation must be equal to zero.

The discriminant of the quadratic equation 25x^2 - 20ax - (y^2 - 4a^2) is given by b^2 - 4ac, where a = 25, b = -20a, and c = -(y^2 - 4a^2). Evaluating the discriminant, we have:

(-20a)^2 - 4(25)(-(y^2 - 4a^2)) = 400a^2 + 100(y^2 - 4a^2) = 400a^2 + 100y^2 - 400a^2 = 100y^2

Since the discriminant is equal to zero, the quadratic equation has a unique solution only when 100y^2 = 0. This implies that y must be equal to zero.

To summarize, for the given system of equations to have exactly one solution, we need the following conditions to be satisfied:

1. The expression inside the square root in Equation 1, y^2-4y+4, must be equal to zero. 2. The value of y must be zero.

Now, let's find the negative values of parameter a that satisfy these conditions.

1. Condition 1: The expression inside the square root in Equation 1, y^2-4y+4, must be equal to zero.

To find the values of y that satisfy this condition, we can solve the quadratic equation y^2-4y+4 = 0. Factoring the quadratic equation, we have:

(y-2)^2 = 0

This equation has a double root at y = 2. Therefore, the expression inside the square root in Equation 1 is equal to zero when y = 2.

2. Condition 2: The value of y must be zero.

Combining both conditions, we find that the system of equations has exactly one solution when y = 2 and a < 0.

Therefore, the negative values of parameter a that satisfy the conditions are a < 0.

To summarize, the negative values of parameter a for which the system of equations has exactly one solution are a < 0.