Pretty basic problem

**mgrade2000** · March 28th 2010, 12:40 PM

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.

**integral** · March 28th 2010, 12:44 PM

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

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

Note:

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

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

**mgrade2000** · March 28th 2010, 12:54 PM

it is .01 not .1 in the original problem

**HallsofIvy** · March 28th 2010, 01:02 PM

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 $.01- .4x+ 4x^2$ .

**mgrade2000** · March 28th 2010, 01:05 PM

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

**e^(i*pi)** · March 28th 2010, 01:12 PM

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 $x \approx 4.48 \times 10^{-3}$

**mgrade2000** · March 28th 2010, 06:28 PM

thanks a lot

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