- Thread starter
- #1

(0.149/(18 - 0.1x - 0.05n)^2) - (44.5/x^2)

The task is to find the variable x

Thanks a lot, your help will mean a lot to me.

- Thread starter PaulJones
- Start date

- Thread starter
- #1

(0.149/(18 - 0.1x - 0.05n)^2) - (44.5/x^2)

The task is to find the variable x

Thanks a lot, your help will mean a lot to me.

- Mar 1, 2012

- 663

does $\dfrac{0.149}{(18 - 0.1x - 0.05n)^2} - \dfrac{44.5}{x^2} = \text{ anything ???}$

- Thread starter
- #3

I'm so so so sorry for the mistake, the above expression is equal to 0.

does $\dfrac{0.149}{(18 - 0.1x - 0.05n)^2} - \dfrac{44.5}{x^2} = \text{ anything ???}$

- Mar 1, 2012

- 663

replacing the constants with $a,b,c,d, e$ to make the algebra easier to follow ...

$\dfrac{a}{(b - cx - dn)^2} - \dfrac{e}{x^2} = 0$

$\dfrac{a}{(b - cx - dn)^2} = \dfrac{e}{x^2}$

$\dfrac{\sqrt{a}}{|b-cx-dn|} = \dfrac{\sqrt{e}}{|x|}$

assuming both $(b-cx-dn)$ and $x$ are same-signed (both positive or both negative) ...

$\dfrac{\sqrt{a}}{b-cx-dn} = \dfrac{\sqrt{e}}{x}$

$x\sqrt{\dfrac{a}{e}} = b-cx-dn$

$x\sqrt{\dfrac{a}{e}}+cx = b-dn$

$x\left(\sqrt{\dfrac{a}{e}} + c \right) = b-dn$

$x = \dfrac{b-dn}{\sqrt{\dfrac{a}{e}} + c}$

assuming $(b-cx-dn)$ and $x$ are different signed (one positive, the other negative) ...

$\dfrac{\sqrt{a}}{dn+cx-b} = \dfrac{\sqrt{e}}{x}$

$x\sqrt{\dfrac{a}{e}} = dn+cx-b$

$x\sqrt{\dfrac{a}{e}}-cx = dn-b$

$x\left(\sqrt{\dfrac{a}{e}} - c \right) = dn-b$

$x = \dfrac{dn-b}{\sqrt{\dfrac{a}{e}} - c}$

Hope this works for you ... if I erred somewhere, I'm sure someone will jump on this thread and point out the mistake.