Solve for x-- the inequality of quadratic

**sumedh** · August 26th 2011, 07:19 AM

Solve $\frac{2x}{x^2-9}\le\frac{1}{x+2}$

------

$x^2-9\not=0$
.'. $x\in R-\{-3,3\}$
and

$x+2\not=0$
.'. $x\in R-\{-2\}$

then converting the original inequality to
$(2x)(x+2)\le(x^2-9)$
$(2x^2+4x)(-x^2+9)\le 0$
$(x^2+4x+9)\le 0$
As Discriminant <0 it has no real roots
so how to do further...

My assumptions:-
$x^2-9$ should not be zero
$x+2$ should also not be zero

Are my assumptions right?
Any help will be highly appreciated.

**anonimnystefy** · August 26th 2011, 07:54 AM

hi sumedh

your "conversion" isn't correct.you cannot multiply an inequality by something you don't know the sign of.

$\frac{2x}{x^2-9}\leq\frac{1}{x+2}$

$\frac{2x}{(x-3)(x+3)}-\frac{1}{x+2}\leq 0$

$\frac{2x(x+2)-(x-3)(x+3)}{(x-3)(x+2)(x+3)}\leq 0$

$\frac{2x^2+4x-(x^2-9)}{(x-3)(x+2)(x+3)}\leq 0$

$\frac{2x^2+4x-x^2+9}{(x-3)(x+2)(x+3)}\leq 0$

$\frac{x^2+4x+9}{(x-3)(x+2)(x+3)}\leq 0$

can you do it from here?

**sumedh** · August 26th 2011, 08:46 AM

i am really very thankful to you

it will be your greatness if you tell few next steps .

**Siron** · August 26th 2011, 08:51 AM

If you know:
$\frac{x^2+4x+9}{(x-3)\cdot (x+3)\cdot (x+2)}\leq 0$
That means:
$x^2+4x+9 \leq 0$
Now solve $x^2+4x+9=0$ .
What do you notice?

After this you only have to make the sign-table to determine where $f(x)\leq 0$

**anonimnystefy** · August 26th 2011, 09:03 AM

hi Siron

actually he should make a sign table for:
$\frac{x^2+4x+9}{(x-3)(x+2)(x+3)}$

because (x-3)(x+2)(x+3) might be negative and thus change the sign of the inequality.so we mustn't multiply by (x-3)(x+2)(x+3).

**Siron** · August 26th 2011, 09:20 AM

Yes, I ment that but maybe I had better said:
$f(x)=\frac{x^2+4x+9}{(x-3)(x+2)(x+3)}\leq 0$

**sumedh** · August 26th 2011, 10:29 PM

x^2+4x+9=0 has imaginary roots.
so it will be always positive
and we will have to consider the denominator now, right??

**Siron** · August 27th 2011, 12:04 AM

Indeed $x^2+4x+9>0, \forall x \in \mathbb{R}$ , an easy way to determine this statement is by using completing the square:
$x^2+4x+9>0 \Leftrightarrow (x^2+4x+4)+5>0 \Leftrightarrow (x+2)^2+5>0 \Leftrightarrow (x+2)^2>-5$
Which is always true!

Go on with the denominator.

**sumedh** · August 27th 2011, 12:36 AM

thanks i got it

**Siron** · August 27th 2011, 12:45 AM

Okay, you're welcome and good work!

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