Inequality solution

**donnagirl** · March 22nd 2010, 11:57 PM

Here's the quadratic I need to solve: $2x^2 -7x > 4$
Now I know the solutions are $4, -\frac {1}{2}$ but my book shows a number line that essentially has $x > 4$ and $x < -\frac {1}{2}$ Why is this the case? Why aren't they all "greater than" inequalities?

**mr fantastic** · March 23rd 2010, 12:05 AM

Originally Posted by donnagirl

Here's the quadratic I need to solve: $2x^2 -7x > 4$
Now I know the solutions are $4, -\frac {1}{2}$ but my book shows a number line that essentially has $x > 4$ and $x < -\frac {1}{2}$ Why is this the case? Why aren't they all "greater than" inequalities?

The inequality is equivalent to $2x^2 - 7x - 4 > 0$ .

Draw a graph of $y = 2x^2 - 7x - 4$ and be sure to label the x-intercepts. For what values of x is y > 0 ....?

**donnagirl** · March 23rd 2010, 12:22 AM

But all y values are greater than 0 (where they are not is part of the interval I don't need..)

**mr fantastic** · March 23rd 2010, 12:24 AM

Originally Posted by donnagirl

But all y values are greater than 0 (where they are not is part of the interval I don't need..)

Draw the graph!!

**donnagirl** · March 23rd 2010, 12:32 AM

I DID draw the graph, it does work--if $x < -\frac{1}{2}$ all values of y are positive. I don't need the part of the interval where the y values are negative. BUT, why can't I manipulate it algebraically? Was I supposed to multiply by a negative or something??

**Bacterius** · March 23rd 2010, 12:41 AM

Consider your quadratic equation :

$2x^2 - 7x - 4$

This can be factorized as :

$(2x + 1)(x - 4)$

Therefore your inequality becomes :

$(2x + 1)(x - 4) > 0$

In order for this inequality to hold, the factors need to have the same sign. Can you follow on ?

**donnagirl** · March 23rd 2010, 12:44 AM

Originally Posted by Bacterius

...
In order for this inequality to hold, the factors need to have the same sign. Can you follow on ?

Not sure what you mean?? That's the same setup I have..yet I got $x > 4, x > -\frac{1}{2}$ . Why is it failing algebraically?

**Bacterius** · March 23rd 2010, 01:07 AM

It is not failing, you are just not going to the end of things. I would draw a sign table :

$\begin{array}{|l|l|} \hline x & -\infty \ \ \ \frac{-1}{2} \ \ \ \ \ \ 4 \ \ \ +\infty \\ \hline 2x + 1 & \ \ - \ \ \ \ \ 0 \ \ + \ \ | \ \ \ \ + \ \ \ \ \ \\ \hline x - 4 & \ \ - \ \ \ \ \ | \ \ \ - \ \ 0 \ \ \ + \ \ \ \ \\ \hline (2x + 1)(x - 4) & \ \ + \ \ \ \ \ 0 \ \ - \ \ 0 \ \ \ + \ \ \ \ \\ \hline \end{array}$

And you get the correct answer $x < \frac{-1}{2}$ and $x > 4$ .

This is how I was taught to do it, makes it a lot easier to visualize

That table was a pain to build.

*PS* : you can achieve the same result using algebra but I forgot how

**donnagirl** · March 23rd 2010, 01:17 AM

Now THAT makes sense! Thanks for your hard work on it Bacterius

I appreciate it!

**Bacterius** · March 23rd 2010, 01:19 AM

Originally Posted by donnagirl

Now THAT makes sense! Thanks for your hard work on it Bacterius

I appreciate it!

It was my pleasure

**mr fantastic** · March 23rd 2010, 02:27 AM

Originally Posted by donnagirl

Now THAT makes sense! Thanks for your hard work on it Bacterius

I appreciate it!

If you had said in the first place that you wanted an algebraic approach rather than a graphical approach (the latter is superior in my opinion, by the way) less time and effort would have been wasted here.

**Krizalid** · March 23rd 2010, 06:22 AM

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