Question

А)-2х^2+5х-27=0. б) х ^2-11х-12=0​в) х^2-5х+6=0г) х^2+5х+6=0​

Answer 1

Вот ответы. Надеюсь, поможет :D

Answer 2

Let's solve each of the quadratic equations:

a) -2x^2 + 5x - 27 = 0

To solve this equation, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = -2, b = 5, and c = -27. Plugging in these values, we get:

x = (-5 ± √(5^2 - 4(-2)(-27))) / (2(-2)) x = (-5 ± √(25 - 216)) / (-4) x = (-5 ± √(-191)) / (-4)

Since the discriminant (b^2 - 4ac) is negative, the equation has no real solutions. Therefore, the equation -2x^2 + 5x - 27 = 0 has no real roots.

b) x^2 - 11x - 12 = 0

Again, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

Here, a = 1, b = -11, and c = -12. Plugging in these values, we get:

x = (-(-11) ± √((-11)^2 - 4(1)(-12))) / (2(1)) x = (11 ± √(121 + 48)) / 2 x = (11 ± √169) / 2 x = (11 ± 13) / 2

So we have two possible solutions: x = (11 + 13) / 2 = 24 / 2 = 12 x = (11 - 13) / 2 = -2 / 2 = -1

Therefore, the equation x^2 - 11x - 12 = 0 has two real solutions: x = 12 and x = -1.

c) x^2 - 5x + 6 = 0

Using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this equation, a = 1, b = -5, and c = 6. Plugging in these values, we get:

x = (-(-5) ± √((-5)^2 - 4(1)(6))) / (2(1)) x = (5 ± √(25 - 24)) / 2 x = (5 ± √1) / 2 x = (5 ± 1) / 2

The solutions are: x = (5 + 1) / 2 = 6 / 2 = 3 x = (5 - 1) / 2 = 4 / 2 = 2

So the equation x^2 - 5x + 6 = 0 has two real solutions: x = 3 and x = 2.

d) x^2 + 5x + 6 = 0

Again, using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

Here, a = 1, b = 5, and c = 6. Plugging in these values, we get:

x = (-5 ± √(5^2 - 4(1)(6))) / (2(1)) x = (-5 ± √(25 - 24)) / 2 x = (-5 ± √1) / 2 x = (-5 ± 1) / 2

The solutions are: x = (-5 + 1) / 2 = -4 / 2 = -2 x = (-5 - 1) / 2 = -6 / 2 = -3

Therefore, the equation x^2 + 5x + 6 = 0 has two real solutions: x = -2 and x = -3.