Question

How do you solve `3 y ^2 - 48 = 0`?

Answer 1

`y = +-4`

Explanation:

We see that there are only two terms in the given problem. One contains the unknown `y` and other is a constant.
`3y^2- 48=0`

Step 1. Keepning the term which contains `y` on the LHS, move the contant term to RHS of the equation. We get

`3y^2= 48`

Step 2. Divide both sides by the common facotor `3`.
`(3y^2)/3 = 48/3`
Simplify

`y^2 = 16`

Step 3. To solve for `y`, Take square root of both sides
`sqrt(y^2) = sqrt16`
Simplify and remember to place `=+-` sign in front of one of the two terms. Selected RHS

`y = +-4`

Answer 2

`y=4, -4`

Explanation:

`3y^2-48=0`

`3` is a common factor between the two terms.
Take `3` and write the remaining terms in brackets:

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

Now, the product of `3` and `(y^2-16)` is `0`.

This means that `color(red)("atleast ONE of the two terms is " 0)`.

`=> 3=0 " or " y^2-16=0`

Obviously, `3!=0`

Then, `y^2-16=0`

Or `y^2-4^2=0`

This is in the form, `color(red)(a^2-b^2)`

And we know that, `color(red)(a^2-b^2=(a-b)(a+b))`

`=>y^2-4^2=0`

`=>(y-4)(y+4)=0`

Again, `color(red)("atleast ONE of the two terms is " 0)`.

`=>y-4=0 " or "y+4=0`

`=>y=4 " or "y=-4`