# A Quick Question of Inequalities

• Jan 3rd 2012, 02:59 AM
dementedsquirrel
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.
• Jan 3rd 2012, 03:03 AM
skeeter
Re: A Quick Question of Inequalities
Quote:

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.
• Jan 3rd 2012, 03:07 AM
dementedsquirrel
Re: A Quick Question of Inequalities
Thanks a lot for your help.
I drew a complete blank for a while.
• Jan 3rd 2012, 03:49 AM
Prove It
Re: A Quick Question of Inequalities
Quote:

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.

\displaystyle \displaystyle \begin{align*} 2x^2 - 7x + 3 &> 0 \\ 2\left(x^2 - \frac{7}{2}x + \frac{3}{2}\right) &> 0 \\ 2\left[x^2 - \frac{7}{2} + \left(-\frac{7}{4}\right)^2 - \left(-\frac{7}{4}\right)^2 + \frac{3}{2}\right] &> 0 \\ 2\left[\left(x - \frac{7}{4}\right)^2 - \frac{49}{16} + \frac{24}{16}\right] &> 0 \\ 2\left[\left(x - \frac{7}{4}\right)^2 - \frac{25}{16}\right] &> 0 \\ \left(x - \frac{7}{4}\right)^2 - \frac{25}{16} &> 0 \\ \left(x - \frac{7}{4}\right)^2 &> \frac{25}{16} \\ \left|x - \frac{7}{4}\right| &> \frac{5}{4} \\ x - \frac{7}{4} < -\frac{5}{4} \textrm{ or }x - \frac{7}{4} &> \frac{5}{4} \\ x < \frac{1}{2} \textrm{ or } x &> 3 \end{align*}
• Jan 3rd 2012, 06:07 AM
dementedsquirrel
Re: A Quick Question of Inequalities
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! :)