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Travka [436]
3 years ago
15

In Algebra need help answering this question

Mathematics
1 answer:
egoroff_w [7]3 years ago
4 0

Answer:

y = 0, y = \frac{1}{3} , y = 7

Step-by-step explanation:

Factor out y^{\frac{3}{5} } from each term

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

y^{\frac{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^{\frac{3}{5} } (y - 7)(3y - 1) = 0

Equate each factor to zero and solve for y

y^{\frac{3}{5} } = 0 ⇒ y = 0

3y - 1 = 0 ⇒ 3y = 1 ⇒ y = \frac{1}{3}

y - 7 = 0 ⇒ y = 7

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