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

2 Answers
May 20, 2016

Answer:

#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#

May 20, 2016

Answer:

#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#