Question

How do you solve `8y^4 - 4y^2 = 0`?

Answer 1

`y= 0" , " y=+-sqrt(2)/2`

Explanation:

Factor out `y^2` giving:`" "y^2(8y^2-4)=0`

So `y^2=0=>y=0`

Or `color(brown)((8y^2-4)=0)color(green)(" " =>" " y^2=1/2)color(purple)(" " =>" " y=+-1/sqrt(2)=+-sqrt(2)/2)`

Answer 2

`y = 0 or y = +-sqrt(1/2)`

Explanation:

Find the factors first., look for a common factor before doing anything else.

`8y^4 - 4y^2 = 0`

`4y^2(2y^2 - 1)= 0`

Either of the factors could be equal to 0.
Make two equations and solve each.

`if 4y^2 = 0" "rArr y^2 = 0 " "rArr y = 0`

if `2y^2 - 1 =0 " "rArr2y^2 = 1" "rArr y^2 = 1/2`

`y = 0 or y = +-sqrt(1/2)`

Answer 3

`y=-sqrt2/2," "0," "sqrt2/2`

Explanation:

First, notice that both terms have a common factor of `4y^2`. This can be factored out of both terms.

`8y^4-4y^2=0`

`4y^2((8y^4)/(4y^2)-(4y^2)/(4y^2))=0`

`4y^2(2y^2-1)=0`

We now have two different terms being multiplied to equal `0`. The first is just `4y^2`, and the other is the entirety of `(2y^2-1)`.

We can use the "Zero Factor Property" here, which basically states that if factors are being multiplied to equal `0`, it stands to reason that either factor must also equal `0`. `(`If `ab=0`, then either `a=0` or `b=0.)`

So here, we know that:

`{(4y^2=0),(2y^2-1=0):}`

Solving the first:

`4y^2=0`

This boils down to our first solution:

`barul|color(white)(a/a)y=0color(white)(a/a)|`

Solving the second:

`2y^2-1=0`

`2y^2=1`

`y^2=1/2`

Taking the square root of both sides (remember to allow positive and negative solutions):

`y=+-sqrt(1/2)=+-1/sqrt2=+-sqrt2/2`

Giving `2` more solutions:

`barul|color(white)(a/a)y=sqrt2/2color(white)(a/a)|`

`barul|color(white)(a/a)y=-sqrt2/2color(white)(a/a)|`