# Math Help - Pretty basic problem

1. ## Pretty basic problem

the problem is $.055=(.0100-2x)^2/x^2$. This becomes $.055=(.0001-4x^2)/x^2$. I multiplied both sides by $x^2$ to get $.055x^2=.0001-4x^2$, then added $4x^2$ to each side to get $4.055x^2=.0001$, divided both sides by 4.055 to get $x^2=2.466*10^-5$ and took the square root of both sides to get x=.004965 or $5.0*10^-3$. My book however says the answer is $4.4*10^-3$ so what have I done wrong? This is actually a chemistry problem so 4.9 to 4.4 is a bigger difference than I am comfortable with.

2. $.055=\frac{(.100-2x)^2}{x^2}$

$.055=\frac{(.01-2x)^2}{x^2}$

Note:

$.1^2=.01\neq.0001$

that is your mistake.
Also, (a+b)^2 = $a^2+2ab+b^2$

3. it is .01 not .1 in the original problem

4. Actually, you are both wrong. $(a+ b)^2$ is NOT equal to $a^2+ b^2$, it is equal to $a^2+ 2ab+ b^2$. $(.1- 2x)^2$ is NOT equal to $.01- 4x^2$, it is equal to $.01- .4x+ 4x^2$.

5. There is no .1!! the original problem is $.055=(0.01-2x)^2/x^2$
There is no .1

6. Originally Posted by mgrade2000
There is no .1!! the original problem is $.055=(0.01-2x)^2/x^2$
There is no .1
$0.055x^2 = \left(10^{-2} - 2x \right)^2 = 10^{-4} - 4 \times 10^{-2}x + 4x^2$

$3.945x^2- 4 \times 10^{-2}x + 10^{-4} = 0$

I suggest using the quadratic formula to solve this one

I get the roots to be $5.66 \times 10^{-3}$ and $x \approx 4.48 \times 10^{-3}$

7. thanks a lot