1. ## Inequality solution

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

2. Originally Posted by donnagirl
Here's the quadratic I need to solve: $\displaystyle 2x^2 -7x > 4$
Now I know the solutions are $\displaystyle 4, -\frac {1}{2}$ but my book shows a number line that essentially has $\displaystyle x > 4$ and $\displaystyle x < -\frac {1}{2}$ Why is this the case? Why aren't they all "greater than" inequalities?
The inequality is equivalent to $\displaystyle 2x^2 - 7x - 4 > 0$.

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

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

4. 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!!

5. I DID draw the graph, it does work--if $\displaystyle x < -\frac{1}{2}$ 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??

$\displaystyle 2x^2 - 7x - 4$

This can be factorized as :

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

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

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

7. 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 $\displaystyle x > 4, x > -\frac{1}{2}$. Why is it failing algebraically?

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

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

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

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

9. Now THAT makes sense! Thanks for your hard work on it Bacterius I appreciate it!

10. Originally Posted by donnagirl
Now THAT makes sense! Thanks for your hard work on it Bacterius I appreciate it!
It was my pleasure

11. 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.

12. first link in my signature.