# Math Help - A silly problem...

1. ## A silly problem...

I know this is a simple problem--I've already done 500 like it. I just can't seem to get the answer.

The question asks you to solve inequalities.

((x-5)/2))^2 < (x-4)/8

Any help would be much appreciated. I tried to set it equal to zero...
I just can't figure it out.

Thanks

2. Originally Posted by PakalCafe
I know this is a simple problem--I've already done 500 like it. I just can't seem to get the answer.

The question asks you to solve inequalities.

((x-5)/2))^2 < (x-4)/8

Any help would be much appreciated. I tried to set it equal to zero...
I just can't figure it out.

Thanks
$(\frac{x-5}{2})^2 < \frac{x-4}{8}$ Multiply by 4 to get:

$(x-5)^2 < \frac{x-4}{2} \Rightarrow x^2 -10x + 25 < \frac{x-4}{2}$ Multiply by 2 again to get:

$2x^2 -10x + 25 -x + 4 < 0$

Can you finish now?

3. Originally Posted by Defunkt
$(\frac{x-5}{2})^2 < \frac{x-4}{8}$ Multiply by 4 to get:

$(x-5)^2 < \frac{x-4}{2} \Rightarrow x^2 -10x + 25 < \frac{x-4}{2}$ Multiply by 2 again to get:

$2x^2 -10x + 25 -x + 4 < 0$

Can you finish now?
It's the WHOLE thing times two though...
that is, 2x^2-20x+50-x+4 < 0 and that's where I got to.
or am I wrong?

4. ## oh wait a second

i figured it out
it [I]is[I] 2x^2-20x+50-X+4<0
Then you factor it out and get
(2x-9)(x-6)<0
Then you get
x<9/2 or x<6

thanks

5. oops, yes, forgot to multiply the rest of the left side.