# Another inequality problem

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• Jun 3rd 2009, 08:09 AM
a69356
Another inequality problem
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
• Jun 3rd 2009, 08:24 AM
Plato
Quote:

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 $5x^2-9x-18 <0$.
• Jun 3rd 2009, 08:26 AM
Rapha
Hi buddy,
Quote:

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