# Math Help - Quadratic inequality confusion

Ok so here goes,

Find the set of values for which the graph of $y=2x^{2}+x-10$ is above the x-axis.

Well that gives us the inequality
$2x^{2}+x-10>0$
$\implies {(2x+5)(x-2)>0}$
$\implies x<\frac {-5}{2}, x>2$

Here is where the confusion is; how do i write the set values because there is an infitesmal amount? I wrote it:

$x=3,4,5,6,7,8........ \infty$
$
x=-3,-4,-5,-6,-7,-8........ -\infty
$

However is this correct?

Sean

2. Consider the case when

$2x+5>0\,\wedge\,x-2>0$ & $2x+5<0\,\wedge\,x-2<0$

3. I dont quite understand, but if
$x-2>0$
$2x+5<0$
The same as in my first post is true?

4. Hello, Sean!

Krizalid is absolutely correct . . .

Find the set of values for which the graph of $y\:=\:2x^2+x-10$ is above the x-axis.

Well, that gives us the inequality: . $2x^{2}+x-10\:>\:0\quad\Rightarrow\quad(2x+5)(x-2)>0$ . Right!

If the product of those two factors is positive:
. . (1) both factors are positive, or
. . (2) both factors are negative.

Case (1): . $\begin{array}{ccccccc}2x+5 & > & 0 & \Rightarrow & x & > & \text{-}\frac{5}{2} \\
x - 2 & > & 0 & \Rightarrow & x & > & 2\end{array}$
. The "stronger" inequality is: . $x \:>\:2$

Case (2): . $\begin{array}{ccccccc}2x + 5 & < & 0 & \Rightarrow & x & < &\text{-}\frac{5}{2} \\
x - 2 & < & 0 & \Rightarrow & x & < & 2\end{array}$
. The "stronger" inequality is: . $x \:<\:\text{-}\frac{5}{2}$

Therefore, the graph is above the x-axis on the intervals: . $\left(\text{-}\infty,\,\text{-}\frac{5}{2}\right),\;(2,\,\infty)$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Of course, if we were allowed to graph the function,
. . we could have "eyeballed" the answers.
Code:
                  |
*            |        *
|
*           |       *
*          |      *
-----o--------+----o-----
-2½ *     | *  2
* |
|

5. Let $f:\mathbf{R}\to\mathbf{R}, \ f(x)=ax^2+bx+c$ a quadratic function, with real roots $x_1,x_2, \ x_1.
Then:
a) If $a>0$ then
$f(x)>0, \ \forall x\in(-\infty,x_1)\cup(x_2,\infty)$
$f(x)<0, \ \forall x\in(x_1,x_2)$

b) If $a<0$ then
$f(x)<0, \ \forall x\in(-\infty,x_1)\cup(x_2,\infty)$
$f(x)>0, \ \forall x\in(x_1,x_2)$

6. Thanks guys,

BTW i "eyeballed" the answer by drawing a quick sketch

7. Originally Posted by Sean12345
BTW i "eyeballed" the answer by drawing a quick sketch
no shame in that