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

3 Answers
Jul 30, 2016

Answer:

#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)#

Jul 30, 2016

Answer:

#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)#

Jul 30, 2016

Answer:

#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)|#