Question

In Algebra need help answering this question

Answer 1

y = 0, y =
`(1)/(3)` , y = 7

Explanation:

Factor out
`y^{(3)/(5) }` from each term

`y^{(3)/(5) }` ( 3
`y^{(10)/(5) }` - 22
`y^{(5)/(5) }` + 7) = 0 , that is

`y^{(3)/(5) }` (3y² - 22y + 7) = 0 ← factor the quadratic

Consider the factors of the product of the coefficient of the y² term and the constant term which sum to give the coefficient of the y- term

product = 3 × 7 = 21 and sum = - 22

The factors are - 21 and - 1

Use these factors to split the y- term

3y² - 21y - y + 7 ( factor the first/second and third/fourth terms )

3y(y - 7) - 1 (y - 7) ← factor out (y - 7) from each term

(y - 7)(3y - 1)

Then

`y^{(3)/(5) }` (y - 7)(3y - 1) = 0

Equate each factor to zero and solve for y

`y^{(3)/(5) }` = 0 ⇒ y = 0

3y - 1 = 0 ⇒ 3y = 1 ⇒ y =
`(1)/(3)`

y - 7 = 0 ⇒ y = 7