Question

6-y-6y2+y3=0 Хелп ми

Answer 1

Ответ:

Объяснение:

6-y-6y²+y³=0

(6-y)-y^2(6-y)=0

(6-y)(1-y²)=0

(6-y)(1-y)(1+y)=0

6-y=0   ,   1-y=0       ,   1+y=0

y1=-6          y2=1            y3=-1

Answer 2

Sure, I can help you with that equation. The equation you provided is a polynomial equation in terms of y. To solve it, we can factor out the common factor, which is y:

y(6 - 6y + y^2) = 0

Now we have a product of two factors that equals zero. According to the zero product property, for the product to be zero, at least one of the factors must be zero. So we have two possibilities:

1. y = 0
2. 6 - 6y + y^2 = 0

Let's solve the second equation:

To solve the quadratic equation 6 - 6y + y^2 = 0, we can try factoring it or using the quadratic formula. In this case, factoring is not straightforward, so let's use the quadratic formula:

y = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 1, b = -6, and c = 6. Substituting these values into the quadratic formula:

y = (-(-6) ± √((-6)^2 - 4(1)(6))) / (2(1)) = (6 ± √(36 - 24)) / 2 = (6 ± √12) / 2 = (6 ± 2√3) / 2

Simplifying further:

y = (6 + 2√3) / 2 = 3 + √3 y = (6 - 2√3) / 2 = 3 - √3

Therefore, the solutions to the equation 6 - 6y + y^2 = 0 are y = 3 + √3 and y = 3 - √3, in addition to the solution y = 0.

To summarize, the solutions to the original equation 6y - 6y^2 + y^3 = 0 are:

y = 0, y = 3 + √3, y = 3 - √3.