Question

Помогите пожалуйста 1) х(2) + 8x + 7 =0 2) x(2) - 19x + 18 =0 3) x(2) - 9x - 10 =0 4) x(2) + 8x - 9 =0 5) x(2) + 5x - 14 =0 6) x(2) - 7x - 30 =0 7) x(2) + 26x + 105 =0 8) x(2) + 35x - 114 =0 ВОТ ПОЖАЛУЙСТА СРОЧНО

Answer 1

1) x 1 = -1
x 2 = -7
2) x 1 = 18
x 2 = 1
3) x 1 = -1
x2 = 10
4) x 1 = -9
x 2 = 1
5) x1 = -7
x2= 2
6) x1 = 10
x2= -3
7) x1 = -21
x2 = -5
8) x1 = -38
x2= 3

Answer 2

1) To solve the equation x^2 + 8x + 7 = 0, we can use the quadratic formula. The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x can be found using the formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 1, b = 8, and c = 7. Plugging these values into the quadratic formula, we get:

x = (-8 ± √(8^2 - 4(1)(7))) / (2(1))

Simplifying further:

x = (-8 ± √(64 - 28)) / 2

x = (-8 ± √36) / 2

x = (-8 ± 6) / 2

This gives us two possible solutions:

x1 = (-8 + 6) / 2 = -1 x2 = (-8 - 6) / 2 = -7

Therefore, the solutions to the equation x^2 + 8x + 7 = 0 are x = -1 and x = -7.

2) To solve the equation x^2 - 19x + 18 = 0, we can again use the quadratic formula. In this case, a = 1, b = -19, and c = 18. Plugging these values into the quadratic formula, we get:

x = (19 ± √((-19)^2 - 4(1)(18))) / (2(1))

Simplifying further:

x = (19 ± √(361 - 72)) / 2

x = (19 ± √289) / 2

x = (19 ± 17) / 2

This gives us two possible solutions:

x1 = (19 + 17) / 2 = 18 x2 = (19 - 17) / 2 = 1

Therefore, the solutions to the equation x^2 - 19x + 18 = 0 are x = 18 and x = 1.

3) To solve the equation x^2 - 9x - 10 = 0, we can once again use the quadratic formula. In this case, a = 1, b = -9, and c = -10. Plugging these values into the quadratic formula, we get:

x = (9 ± √((-9)^2 - 4(1)(-10))) / (2(1))

Simplifying further:

x = (9 ± √(81 + 40)) / 2

x = (9 ± √121) / 2

x = (9 ± 11) / 2

This gives us two possible solutions:

x1 = (9 + 11) / 2 = 10 x2 = (9 - 11) / 2 = -1

Therefore, the solutions to the equation x^2 - 9x - 10 = 0 are x = 10 and x = -1.

4) To solve the equation x^2 + 8x - 9 = 0, we can once again use the quadratic formula. In this case, a = 1, b = 8, and c = -9. Plugging these values into the quadratic formula, we get:

x = (-8 ± √(8^2 - 4(1)(-9))) / (2(1))

Simplifying further:

x = (-8 ± √(64 + 36)) / 2

x = (-8 ± √100) / 2

x = (-8 ± 10) / 2

This gives us two possible solutions:

x1 = (-8 + 10) / 2 = 1 x2 = (-8 - 10) / 2 = -9

Therefore, the solutions to the equation x^2 + 8x - 9 = 0 are x = 1 and x = -9.

5) To solve the equation x^2 + 5x - 14 = 0, we can once again use the quadratic formula. In this case, a = 1, b = 5, and c = -14. Plugging these values into the quadratic formula, we get:

x = (-5 ± √(5^2 - 4(1)(-14))) / (2(1))

Simplifying further:

x = (-5 ± √(25 + 56)) / 2

x = (-5 ± √81) / 2

x = (-5 ± 9) / 2

This gives us two possible solutions:

x1 = (-5 + 9) / 2 = 2 x2 = (-5 - 9) / 2 = -7

Therefore, the solutions to the equation x^2 + 5x - 14 = 0 are x = 2 and x = -7.

6) To solve the equation x^2 - 7x - 30 = 0, we can once again use the quadratic formula. In this case, a = 1, b = -7, and c = -30. Plugging these values into the quadratic formula, we get:

x = (7 ± √((-7)^2 - 4(1)(-30))) / (2(1))

Simplifying further:

x = (7 ± √(49 + 120)) / 2

x = (7 ± √169) / 2

x = (7 ± 13) / 2

This gives us two possible solutions:

x1 = (7 + 13) / 2 = 10 x2 = (7 - 13) / 2 = -3

Therefore, the solutions to the equation x^2 - 7x - 30 = 0 are x = 10 and x = -3.

7) To solve the equation x^2 + 26x + 105 = 0, we can once again use the quadratic formula. In this case, a = 1, b = 26, and c = 105. Plugging these values into the quadratic formula, we get:

x = (-26 ± √(26^2 - 4(1)(105))) / (2(1))

Simplifying further:

x = (-26 ± √(676 - 420)) / 2

x = (-26 ± √256) / 2

x = (-26 ± 16) / 2

This gives us two possible solutions:

x1 = (-26 + 16) / 2 = -5 x2 = (-26 - 16) / 2 = -21

Therefore, the solutions to the equation x^2 + 26x + 105 = 0 are x = -5 and x = -21.

8) To solve the equation x^2 + 35x - 114 = 0, we can once again use the quadratic formula. In this case, a = 1, b = 35, and c = -114. Plugging these values into the quadratic formula, we get:

x = (-35 ± √(35^2 - 4(1)(-114))) / (2(1))

Simplifying further:

x = (-35 ± √(1225 + 456)) / 2

x = (-35 ± √1681) / 2

x = (-35 ± 41) / 2

This gives us two possible solutions:

x1 = (-35 + 41) / 2 = 3 x2 = (-35 - 41) / 2 = -38

Therefore, the solutions to the equation x^2 + 35x - 114 = 0 are x = 3 and x = -38.

I hope this helps! Let me know if you have any further questions.