(1-2y)(1+2y)>=0

Question

I want to know the value of this equation and the procedure of solving eqn

Answer 1

( 1 - 2y)(1+2y)> 0

1-2y>0 and 1+2y>0

-2y> -1 and 2y> -1

y<1/2 and y> - 1/2

y is in interval ( - 1/2, 1/2)

Answer 2

First let's deal with the equals case. This means the solution is y=1/2 or y=-1/2. This will make one or other of the contents of the brackets zero, so the product is zero.

Now deal with the inequality. To satisfy greater than zero, the contents of the brackets must both be positive or both negative. Therefore 1-2y>0 and 1+2y>0, so 2y<1 and 2y>-1. Therefore y<1/2 and y>-1/2 which is written -1/2<y<1/2. But we also know that y=1/2 or y=-1/2 are solutions, therefore we can write -1/2≤y≤1/2.

To satisfy both negative, 1-2y<0, so 2y>1, y>1/2 and 1+2y<0, so y<-1/2. These conditions cannot both be satisfied with the same value of y, therefore the only solution is -1/2≤y≤1/2, that is y is between -1/2 and 1/2 inclusive.