Re: inequation

• To: mathgroup at smc.vnet.net
• Subject: [mg13903] Re: inequation
• From: "Allan Hayes" <hay at haystack.demon.cc.uk>
• Date: Sun, 6 Sep 1998 02:55:40 -0400
• References: <6sil72\$4at@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Client Préferé <@ping.be> wrote in message <6sil72\$4at at smc.vnet.net>...
>resoudre l inequatin
>
>(3-2x/x-1)² <ou=(6-5x/x+2)²
>
>
>merci pour vos solution
>
>                      nicolas d aout
>

Nicolas,

When x = 0 both sides are indeterminate.

Otherwise we have
(3 -2 -1)^3 <= (6 -5 +2)^2
<=>          0^2 <= 3^2
<=>          0 <= 9
<=>          True

Mathematica looks only at the general case

((3-2x/x-1)^2 <=(6-5x/x+2)^2)

True

However, you may have intended ((3-2x/(x-1))^2 <=(6-5x/(x+2))^2) And in
this case we merely get

((3-2x/(x-1))^2 <=(6-5x/(x+2))^2)

2 x   2          5 x  2
(3 - ------)  <= (6 - -----)
-1 + x           2 + x

<<Algebra`InequalitySolve`

(these can be lloked up in the menu Help > Help > Add-ons > Standard
Packages)

and then we get

InequalitySolve[((3-2x/(x-1))^2 <=(6-5x/(x+2))^2),x] Out[9]=
1                                           1 (- (-5 - Sqrt[61]) <= x)
< -2 || (-2 < x) <= - ||
2                                           2

1
x >= - (-5 + Sqrt[61])
2

Of course, this still leaves out the special cases, x=1, x=-2.

Best wishes

------------------------------------------------------------- Allan
Hayes
Mathematica Training and Consulting
Leicester UK
http://www.haystack.demon.co.uk
hay at haystack.demon.co.uk
voice: +44 (0)116 271 4198
fax: +44(0)116 271 8642

```

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