1. ## Another inequality problem

Solve the below inequality:

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

I tried solving this as below :

squaring both side and after solving

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

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

2. Originally Posted by a69356
Solve the below inequality:
$\displaystyle ({1 - \frac {x+2} {x^2}}) ^ {1/2} < \frac 2 3$

I tried solving this as below :

squaring both side and after solving

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

3. Hi buddy,
Originally Posted by a69356
Solve the below inequality:

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

I tried solving this as below :

squaring both side and after solving

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

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

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

Multiply by x^2

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

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

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

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

Beaten to Plato

Yours Rapha