Quadratic inequality confusion

**Sean12345** · October 7th 2007, 09:06 AM

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:

$x=3,4,5,6,7,8........ \infty$
$<br /> x=-3,-4,-5,-6,-7,-8........ -\infty<br />$

However is this correct?

Thanks in advance,

Sean

**Krizalid** · October 7th 2007, 09:10 AM

Consider the case when

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

**Sean12345** · October 7th 2007, 09:39 AM

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

**Soroban** · October 7th 2007, 09:48 AM

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

Case (2): . $\begin{array}{ccccccc}2x + 5 & < & 0 & \Rightarrow & x & < &\text{-}\frac{5}{2} \\<br /> x - 2 & < & 0 & \Rightarrow & x & < & 2\end{array}$ . 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
                * |
                  |

**red_dog** · October 7th 2007, 09:58 AM

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

**Sean12345** · October 7th 2007, 10:00 AM

Thanks guys,

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

**Jhevon** · October 7th 2007, 10:19 AM

Originally Posted by Sean12345

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

no shame in that

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