Question

How do you solve `y^ { 3/ 2} = 4y`?

Answer 1

`y=0color(white)("xxx")orcolor(white)("xxx")y=16`

Explanation:

Given
`color(white)("XXX")y^(3/2)=4y`

After squaring both sides:
`color(white)("XXX")y^3=16y^2`

Subtract `16y^2` from both sides
`color(white)("XXX")y^3-16y^2=0`

Factor the left side
`color(white)("XXX")y^2(y-16)=0`

Which implies either
`color(white)("XXX")y^2=0`
`color(white)("XXX")rarr y=0`
or
`color(white)("XXX")(y-16)=0`
`color(white)("XXX")rarr y=16`

Answer 2

See a solution process below:

Explanation:

First, square both sides of the eqution:

`(y^(3/2))^2 = (4y)^2`

`y^(3/2 xx 2) = 16y^2`

`y^3 = 16y^2`

`y^3 - color(red)(16y^2) = 16y^2 - color(red)(16y^2)`

`y^3 - 16y^2 = 0`

Next, factor the left side of the equation:

`(y^2 * y) - (y^2 * 16) = 0`

`y^2(y - 16) = 0`

Now, solve each term on the left side of the equation for `0`:

Solution 1:

`y^2 = 0`

`sqrt(y^2) = +-sqrt(0)`

`y = +-0`

`y = 0`

Solution 2:

`y - 16 = 0`

`y - 16 + color(red)(16) = 0 + color(red)(16)`

`y - 0 = 16`

`y = 16`

The Solutions Are: `y = 0` and `y = 16`