Another inequality problem

**a69356** · June 3rd 2009, 08:09 AM

Solve the below inequality:

$ ({1 - \frac {x+2} {x^2}}) ^ {1/2} < \frac 2 3 $

I tried solving this as below :

squaring both side and after solving

$ (5x^2-9x+18) < 0 $

and the factors of the above equation comes to be
x= (9 +(81 - 360)^1/2)/10,(9 -(81 - 360)^1/2)/10

which turns out to be complex number

Please let me know how to proceed further.

Any help would be appreciated.

Thanks

**Plato** · June 3rd 2009, 08:24 AM

Originally Posted by a69356

Solve the below inequality:
$ ({1 - \frac {x+2} {x^2}}) ^ {1/2} < \frac 2 3 $

I tried solving this as below :

squaring both side and after solving

$ \color{red}(5x^2-9x+18) < 0$ here is a mistake.

It should be $5x^2-9x-18 <0$ .

**Rapha** · June 3rd 2009, 08:26 AM

Hi buddy,

Originally Posted by a69356

Solve the below inequality:

$ ({1 - \frac {x+2} {x^2}}) ^ {1/2} < \frac 2 3 $

I tried solving this as below :

squaring both side and after solving

$ (5x^2-9x+18) < 0 $

This is already false.

$ ({1 - \frac {x+2} {x^2}}) ^ {1/2} < \frac 2 3 $

squaring leads to

$1 - \frac{x+2}{x^2} < \frac 4 9$

Multiply by x^2

$x^2 - (x+2) < \frac 4 9 x^2$

$x^2 - \frac 4 9 x^2 - (x+2)<0$

$\frac 5 9 x^2 -x-2 < 0$

solve this and you are nearly done. You still have to consider the domain of f(x)

Beaten to Plato

Yours Rapha

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