n-2n^2=0
n(1 - 2n) = 0
n = 0 or 1 - 2n = 0
n = 0 or n = 1/2
How did you get 1 - 2n = 0?
I took out a common factor of n
expand n(1-2n)
What do you get ?
Yes, but how did the first n in n(1 - 2n) go away?
It didn't, n = 0 became a solution.
Are you not familiar that if you have a factored form = 0, such as
(A)(B) = 0 , then
A = 0 or B = 0 ???
To solve the equation n - 2n^2 = 0, we can follow these steps:
Step 1: Rearrange the equation:
n - 2n^2 = 0
Rewrite it in standard form with the term involving n^2 first:
-2n^2 + n = 0
Step 2: Factor out n to solve for n:
n(-2n + 1) = 0
This equation will be true if either n = 0 or -2n + 1 = 0.
Step 3: Solve for n:
If n = 0, then one solution is n = 0.
If -2n + 1 = 0, we can solve for n by isolating n:
-2n = -1
Divide both sides by -2:
n = 1/2
So, the solutions to the equation n - 2n^2 = 0 are n = 0 and n = 1/2.