Question

Помогите разобраться с дискриминанте как решать пожалуйста пару примеров) заранее спасибо)

Answer 1

Например есть пример: 10х^2+15х+5=0 (выглядит как: ах^2+bx+c=0)
формула дискриминанта: D=b^2-4ac
подставляем цифры с примера в формулу:
D=15^2-4*5*10=225-200=25
потом нужно найти корень с . √D=5
Ищем х₁ и х₂:
формулы: х₁=(-b+√D)/2*a    x₂=(-b-√D)/2*a
x₁=(-15+5)/2*10=-1/2=-0.5 x₂=(-15-5)/20=-1

Answer 2

Understanding the Discriminant

The discriminant is a mathematical term used in quadratic equations to determine the nature of the solutions. It helps us understand whether the equation has real solutions, imaginary solutions, or repeated solutions. The discriminant is calculated using the coefficients of the quadratic equation.

The general form of a quadratic equation is: ax^2 + bx + c = 0, where a, b, and c are constants.

The discriminant, denoted as Δ (delta), is calculated using the formula: Δ = b^2 - 4ac.

Now, let's discuss the different cases based on the value of the discriminant.

Case 1: Δ > 0 (Positive Discriminant)

If the discriminant is positive, it means that the quadratic equation has two distinct real solutions. In other words, the equation intersects the x-axis at two different points.

Case 2: Δ = 0 (Zero Discriminant)

If the discriminant is zero, it means that the quadratic equation has one real solution. In other words, the equation intersects the x-axis at a single point. This is also known as a repeated root or a perfect square trinomial.

Case 3: Δ < 0 (Negative Discriminant)

If the discriminant is negative, it means that the quadratic equation has no real solutions. In other words, the equation does not intersect the x-axis. Instead, it has two complex conjugate solutions, which are imaginary numbers.

Solving Quadratic Equations Using the Discriminant

To solve a quadratic equation using the discriminant, follow these steps:

1. Calculate the discriminant using the formula: Δ = b^2 - 4ac. 2. Determine the nature of the solutions based on the value of the discriminant: - If Δ > 0, the equation has two distinct real solutions. - If Δ = 0, the equation has one real solution. - If Δ < 0, the equation has no real solutions. 3. If the equation has real solutions, use the quadratic formula to find the values of x: - x = (-b ± √Δ) / (2a), where ± indicates two possible solutions.

Let's go through a couple of examples to illustrate how to solve quadratic equations using the discriminant.

Example 1:

Consider the quadratic equation: 2x^2 - 5x + 2 = 0.

1. Calculate the discriminant: Δ = (-5)^2 - 4(2)(2) = 25 - 16 = 9. 2. Since Δ > 0, the equation has two distinct real solutions. 3. Use the quadratic formula to find the values of x: - x = (-(-5) ± √9) / (2(2)) = (5 ± 3) / 4. - The two solutions are: x = (5 + 3) / 4 = 2 and x = (5 - 3) / 4 = 1/2.

Therefore, the solutions to the equation 2x^2 - 5x + 2 = 0 are x = 2 and x = 1/2.

Example 2:

Consider the quadratic equation: x^2 + 4x + 5 = 0.

1. Calculate the discriminant: Δ = (4)^2 - 4(1)(5) = 16 - 20 = -4. 2. Since Δ < 0, the equation has no real solutions. 3. The solutions are two complex conjugate numbers. - The solutions can be written as: x = (-4 ± √(-4)) / (2(1)) = (-4 ± 2i) / 2. - Simplifying further, we get: x = -2 ± i.

Therefore, the solutions to the equation x^2 + 4x + 5 = 0 are x = -2 + i and x = -2 - i.

I hope these examples help you understand how to solve quadratic equations using the discriminant. Let me know if you have any further questions!