Question

X²+6x+4=0x²+106x+693=0​

Answer 1

Ответ:

не знаю, нужно ли считать выражение  под корнем, но написала так

Объяснение:

Answer 2

Ответ:

1.x^2+6x+4=0

a=1; b=6; c=4

D=b^2-4ac

D=6^2-4*1*4=36-16=20

D=20.

2.x^2+106x+693=0

a=1; b=106; c=693

D=b^2-4ac

D=106^2-4*1*693=11236-2772=8464

D=8464:

Просьба:

1. Поставь 5 звёзд.

2. Поставь спасибо.

3. Сделай лучшим ответом.

Answer 3

To solve these quadratic equations, you can use the quadratic formula:

For a quadratic equation in the form of ax^2 + bx + c = 0, the quadratic formula is:

x = (-b ± √(b^2 - 4ac)) / (2a)

Let's solve the two equations you provided:

1. x^2 + 6x + 4 = 0 a = 1, b = 6, and c = 4

   x = (-6 ± √(6^2 - 4 * 1 * 4)) / (2 * 1) x = (-6 ± √(36 - 16)) / 2 x = (-6 ± √20) / 2 x = (-6 ± 2√5) / 2

   Now, simplify the expression: x = -3 ± √5

So, the solutions for the first equation are: x₁ = -3 + √5 x₂ = -3 - √5

2. x^2 + 106x + 693 = 0 a = 1, b = 106, and c = 693

   x = (-106 ± √(106^2 - 4 * 1 * 693)) / (2 * 1) x = (-106 ± √(11236 - 2772)) / 2 x = (-106 ± √8464) / 2 x = (-106 ± 92) / 2

Now, simplify the expression: x₁ = (-106 + 92) / 2 = -14 / 2 = -7 x₂ = (-106 - 92) / 2 = -198 / 2 = -99

So, the solutions for the second equation are: x₁ = -7 x₂ = -99

These are the solutions for the given quadratic equations.