Question

Решить уравнение для ОГЭ 9 класс

Answer 1

А где уравнение? На ОГЭ всегда разные дают же .

Answer 2

Solving Equations for the 9th Grade OGE

To solve an equation for the 9th grade OGE (General Education Exam), we need to find the value of the variable that satisfies the equation. Let's look at some examples:

Example 1: (x-3)(x+4) = 0

To solve this equation, we can use the zero product property, which states that if the product of two factors is equal to zero, then at least one of the factors must be zero. So, we set each factor equal to zero and solve for x:

x - 3 = 0 --> x = 3

x + 4 = 0 --> x = -4

Therefore, the solutions to the equation are x = 3 and x = -4.

Example 2: 3x^2 - 4x + 2 = 0

To solve this quadratic equation, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this equation, a = 3, b = -4, and c = 2. Plugging these values into the quadratic formula, we get:

x = (-(-4) ± √((-4)^2 - 4 * 3 * 2)) / (2 * 3)

Simplifying further:

x = (4 ± √(16 - 24)) / 6

x = (4 ± √(-8)) / 6

Since the discriminant (b^2 - 4ac) is negative, the equation has no real solutions. Therefore, there are no solutions to this equation.

Example 3: x^2 - 4x - 5 = 0

Again, we can use the quadratic formula to solve this equation:

x = (-b ± √(b^2 - 4ac)) / (2a)

Here, a = 1, b = -4, and c = -5. Plugging these values into the quadratic formula, we get:

x = (-(-4) ± √((-4)^2 - 4 * 1 * (-5))) / (2 * 1)

Simplifying further:

x = (4 ± √(16 + 20)) / 2

x = (4 ± √36) / 2

x = (4 ± 6) / 2

So, the solutions to the equation are:

x1 = (4 + 6) / 2 = 10 / 2 = 5

x2 = (4 - 6) / 2 = -2 / 2 = -1

Therefore, the solutions to the equation are x = 5 and x = -1.

These are just a few examples of equations that can be solved for the 9th grade OGE. Remember to always check your solutions by substituting them back into the original equation to ensure they are valid.