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

2 Answers
May 20, 2016

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

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