Thread: A Quick Question of Inequalities

I was a bit confused on something.
It's inequalities.

The question is
Find the set values of x for which:
2x^2 - 7x + 3 > 0

So, I did this.
(2x - 1) (x - 3) > 0
And after quickly sketching it onto a graph, chose the 'inside' regions.
x > 1/2
x < 3

My question is, Is it actually supposed to be the 'outside' regions? ie.
x < 1/2
x > 3

Thanks.

Originally Posted by dementedsquirrel

I was a bit confused on something.
It's inequalities.

The question is
Find the set values of x for which:
2x^2 - 7x + 3 > 0

So, I did this.
(2x - 1) (x - 3) > 0
And after quickly sketching it onto a graph, chose the 'inside' regions.
x > 1/2
x < 3

My question is, Is it actually supposed to be the 'outside' regions? ie.
x < 1/2
x > 3

Thanks.

outside ... note the graph of y = 2x^2 - 7x + 3 is a parabola that opens upward and has roots at x = 1/2 and x = 3. the y-values of the parabola are negative (i.e. less than 0) between those two roots.

Thanks a lot for your help.
I drew a complete blank for a while.

Originally Posted by dementedsquirrel

I was a bit confused on something.
It's inequalities.

The question is
Find the set values of x for which:
2x^2 - 7x + 3 > 0

So, I did this.
(2x - 1) (x - 3) > 0
And after quickly sketching it onto a graph, chose the 'inside' regions.
x > 1/2
x < 3

My question is, Is it actually supposed to be the 'outside' regions? ie.
x < 1/2
x > 3

Thanks.

The most direct method to solve quadratic inequalities is to complete the square and make use of the absolute value function.

Thanks.
This way is good, but the best way for me (during an exam - for a couple of marks) personally is to factorise and work it out accordingly.
Though thanks for the tips!