Question

4(4×-11)²-13(4×-11)+3=0 знайти ×¹ і ײ?​

Answer 1

(4х-11)(4(4х-11)-13)+3=0

(4х-11)(16х-44-13)+3=0

(4х-11)(16х-57)+3=0

64х^2-404х+627+3=0

32х^2-202х+315=0

(16х-45)(2х-7)=0

Х1= 45÷16

Х2=7÷2

Answer 2

To solve the quadratic equation 4(4×-11)² - 13(4×-11) + 3 = 0, we can use the substitution method. Let's introduce a new variable, say, "y," and make the substitution:

y = 4×-11

Now, the equation becomes a quadratic equation in terms of "y":

4y² - 13y + 3 = 0

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

y = (-b ± √(b² - 4ac)) / 2a

where a = 4, b = -13, and c = 3.

Now, let's plug in these values:

y = (13 ± √((-13)² - 4 * 4 * 3)) / (2 * 4)

y = (13 ± √(169 - 48)) / 8

y = (13 ± √121) / 8

y = (13 ± 11) / 8

Now, we have two possible values for "y":

1. y₁ = (13 + 11) / 8 = 24 / 8 = 3
2. y₂ = (13 - 11) / 8 = 2 / 8 = 1/4

Now, we need to convert these "y" values back to "x" using the substitution we made earlier:

1. For y = 3:

4x - 11 = 3

4x = 3 + 11

4x = 14

x₁ = 14 / 4

x₁ = 7/2

2. For y = 1/4:

4x - 11 = 1/4

4x = 1/4 + 11

4x = 45/4

x₂ = (45/4) / 4

x₂ = 45/16

Therefore, the solutions for the original equation are:

x₁ = 7/2 (or 3.5)

x₂ = 45/16 (or approximately 2.8125)