Solve for x-- the inequality of quadratic

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

------

$\displaystyle x^2-9\not=0$

.'. $\displaystyle x\in R-\{-3,3\}$

and

$\displaystyle x+2\not=0$

.'. $\displaystyle x\in R-\{-2\}$

then converting the original inequality to

$\displaystyle (2x)(x+2)\le(x^2-9)$

$\displaystyle (2x^2+4x)(-x^2+9)\le 0$

$\displaystyle (x^2+4x+9)\le 0$

As Discriminant <0 it has no real roots

so how to do further...

My assumptions:-

$\displaystyle x^2-9$should not be zero

$\displaystyle x+2$should also not be zero

Are my assumptions right?

Any help will be highly appreciated.

Re: Solve for x-- the inequality of quadratic

hi sumedh

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

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

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

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

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

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

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

can you do it from here?

Re: Solve for x-- the inequality of quadratic

i am really very thankful to you :)

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

Re: Solve for x-- the inequality of quadratic

If you know:

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

That means:

$\displaystyle x^2+4x+9 \leq 0$

Now solve $\displaystyle x^2+4x+9=0$.

What do you notice?

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

Re: Solve for x-- the inequality of quadratic

hi Siron

actually he should make a sign table for:

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

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).

Re: Solve for x-- the inequality of quadratic

Yes, I ment that but maybe I had better said:

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

Re: Solve for x-- the inequality of quadratic

x^2+4x+9=0 has imaginary roots.

so it will be always positive

and we will have to consider the denominator now, right??

Re: Solve for x-- the inequality of quadratic

Indeed $\displaystyle x^2+4x+9>0, \forall x \in \mathbb{R}$, an easy way to determine this statement is by using completing the square:

$\displaystyle x^2+4x+9>0 \Leftrightarrow (x^2+4x+4)+5>0 \Leftrightarrow (x+2)^2+5>0 \Leftrightarrow (x+2)^2>-5$

Which is always true!

Go on with the denominator.

Re: Solve for x-- the inequality of quadratic

Re: Solve for x-- the inequality of quadratic

Okay, you're welcome and good work! :)