Solve for x- the inequality of quadratic

[Sumedh]

Solve for x-- the inequality of quadratic

Homework Statement

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

The Attempt at a Solution

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

 

[micromass]

then converting the original inequality to
$$(2x)(x+2)\le(x^2-9)$$

I see what you did here, but it's not correct. It's not true that

$$\frac{a}{b}\leq \frac{c}{d}$$

implies that $ad\leq bc$. For example

$$\frac{1}{-1}\leq \frac{1}{1}$$

but it's not true that $1\leq -1$. The problem is of course that negative numbers reverse the inequality sign.

How do we do such a thing then?? Well, first we move everything to the left side of the equation:

$$\frac{2x}{x^2-9}-\frac{1}{x+2}\leq 0$$

and now you must try to add these fractions.

 

[Ray Vickson]

Homework Statement

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

The Attempt at a Solution

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

In general, if a <= b we have a*c <= b*c if c > 0 and a*c >= b*c if c < 0. Thus, if x+1 > 0 you have 2x(x+2)/(x^2-9) <= 1 and if x+2 < 0 you have the opposite inequality. Now, in each case, look at whether x^2-9 is positive or negative.

RGV

 

[uart]

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

I often prefer to multiply by a square to avoid splitting into too many +/- cases. It's a compromise though, between getting a higher order system versus less case splitting.

$$\frac{2x (x+2)^2}{x^2-9}\le x+2$$

I'll split into cases for the (x^2-9) factor to keep the order of the system under control.

$$2x (x+2)^2 \le (x+2) (x^2-9)\,\,\,\, : \,\, |x|>3$$
$$2x (x+2)^2 \ge (x+2) (x^2-9)\,\,\,\, : \,\, |x|<3$$

Now we've got a cubic, but one factor is already out, so rearrange without re-absorbing the "outed" factor.

$$(x^2 + 4x + 9)(x+2) \le 0\,\,\,\, : \,\, |x|>3$$
$$(x^2 + 4x + 9)(x+2) \ge 0\,\,\,\, : \,\, |x|<3$$

Here we have a cubic with just one real root at x=-2, so it's pretty easy to deduce where it's greater or less than zero. Just combine those regions (intersection) with the |x|>3 and |x|<3 constraints and make sure you omit the singular points that weren't in the original domain (-2).

 

[Sumedh]

Thank you very much.