Question

5x^2-4x+2=0 7x^2+6x-1=0

Answer 1

Ответ:

1) -2 и 0,4

2) -1  и  $\frac{1}{7}$

Объяснение:

1)5$x^{2}$-4х+2=0

x(5x-4)= -2

x= -2    и    5х=2

                  х=0,4

2) 7$x^{2}$+6х-1=0

по следствию 2 теоремы Виетта:

х= -1

х= $\frac{1}{7}$

Answer 2

Вот..................

Answer 3

It looks like you've provided two quadratic equations. If you're looking to solve these equations for their respective values of x, I can help you with that.

Let's solve each equation separately:

1. Equation 1: 5x^2 - 4x + 2 = 0

We can use the quadratic formula to solve this equation:

The quadratic formula is given by: x = (-b ± √(b^2 - 4ac)) / 2a

For this equation, a = 5, b = -4, and c = 2.

Plugging in these values: x = (-(-4) ± √((-4)^2 - 4 * 5 * 2)) / (2 * 5) x = (4 ± √(16 - 40)) / 10 x = (4 ± √(-24)) / 10

Since the discriminant (b^2 - 4ac) is negative, the solutions will involve complex numbers.

x = (4 ± √24i) / 10 x = (4 ± 2√6i) / 10 x = (2 ± √6*i) / 5

So, the solutions for the first equation are: x = (2 + √6i) / 5 x = (2 - √6i) / 5

2. Equation 2: 7x^2 + 6x - 1 = 0

Again, we can use the quadratic formula:

a = 7, b = 6, and c = -1.

Plugging in these values: x = (-6 ± √(6^2 - 4 * 7 * (-1))) / (2 * 7) x = (-6 ± √(36 + 28)) / 14 x = (-6 ± √64) / 14

x = (-6 ± 8) / 14

The solutions for the second equation are: x = (-6 + 8) / 14 = 2 / 14 = 1/7 x = (-6 - 8) / 14 = -14 / 14 = -1

So, the solutions for the second equation are: x = 1/7 x = -1

Please note that the first equation's solutions involve complex numbers, while the second equation's solutions are real numbers.