Question

2x^2-2x+0,5=0 через дискриминант ​

Answer 1

$2 {x}^{2} - 2x + 0.5 = 0 \: \: \: \: \: |\cdot 10 \\ 20 {x}^{2} - 20x + 5 = 0 \\ D = {( - 20)}^{2} - 4 \cdot20 \cdot5 = 400 - 400 = 0 \\ x = \frac{20}{2 \cdot20} = \frac{1}{2} = 0.5$

Ответ: 0,5

Answer 2

Ответ:

0.5

Пошаговое объяснение:

Д=(-2)^2-4*2*0.5=4-8*0.5=4-4=0

Д=0

х=-b/2a=2/4=0.5

Answer 3

To find the solutions of the quadratic equation $2x^2 - 2x + 0.5 = 0$ using the discriminant, we can use the quadratic formula:

The quadratic formula is given by: $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$,

where $a$, $b$, and $c$ are the coefficients of the quadratic equation $ax^2 + bx + c = 0$.

For our equation, $2x^2 - 2x + 0.5 = 0$, we have:

$a = 2$, $b = -2$, $c = 0.5$.

Now, let's calculate the discriminant, denoted by $D$, which is given by: $D = b^2 - 4ac$.

$D = (-2)^2 - 4 \times 2 \times 0.5 = 4 - 4 = 0$.

Now, depending on the value of the discriminant $D$:

1. If $D > 0$, the quadratic equation has two distinct real solutions.
2. If $D = 0$, the quadratic equation has one real solution (a repeated root).
3. If $D < 0$, the quadratic equation has two complex solutions (conjugate pair of complex numbers).

In our case, since $D = 0$, the quadratic equation has one real solution (a repeated root).

Let's find the value of $x$:

$x = \frac{{-b}}{{2a}} = \frac{{-(-2)}}{{2 \times 2}} = \frac{2}{4} = \frac{1}{2}$.

So, the quadratic equation $2x^2 - 2x + 0.5 = 0$ has one real solution, which is $x = \frac{1}{2}$.