How do you factor 2m^2-m<1?

Question

This problem on my assignment is giving me trouble

Answer 1

Greetings,

2m^2 - m - 1 < 0
2m^2 - 2m + m - 1 < 0
2m(m -1) +1(m - 1)< 0
(2m + 1)(m -1)<0

- where the factor is neagtive
o where the factor is zero
+ where the factor is positive

m -1 ----------o+++++++
2m + 1 --o+++++++++
...........-1/2..1.....

To be negative -1/2< m < 1

Regards

Answer 2

-1/2

Answer 3

Write it 2m^2 - m - 1 <0

Factor it and rewrite it as (2m+1)(m-1) <0.

This happens when EXACTLY ONE -- i.e., one but not both -- of 2m+1 and m-1 is less than 0.

Well, 2m+1 is less than 0 if and only if m < 1/2
And m-1<0 if and only if m < 1

So exactly one of them is less than 0

when and only when

1/2 < m < 1.

Answer 4

2m^2-m<1 => 2m^2-m-1<0 => (m-1)(2m+1)<0 then we suppose that m>0, so 2m+1 will be the bigger one. Since it is the bigger one, then we get 2m+1<0, so m<-1/2

Answer 5

2m^2 - m - 1 < 0
(2m + 1)(m - 1) < 0
-1/2 < m < 1

Answer 6

set the eqn to 0
2m^2-m =0
m(2m-1) =0
therefore ,

m= 1 or 2m-1 =0
i.e m=1 and m=1/2

1/2 <= m >=1