MHB Solve for X | Math Problem | Paul's Question

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Paul is seeking help to solve the equation (0.149/(18 - 0.1x - 0.05n)^2) - (44.5/x^2) = 0 for the variable x. Initially, there was confusion as the expression was presented without an equation. After clarification, the expression was correctly set to equal zero. The solution process involves substituting constants and simplifying the equation, leading to two potential solutions for x based on the signs of the terms involved. The final expressions for x are x = (b - dn) / (√(a/e) + c) and x = (dn - b) / (√(a/e) - c), depending on the sign assumptions.
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Hi, my name is Paul and I'm new to this forum. I'm having a math problem that I'm unable to find a solution for it at all, i have tried many solutions but unable to find the x
(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.
 
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what you have posted is an expression in two variables, not an equation

does $\dfrac{0.149}{(18 - 0.1x - 0.05n)^2} - \dfrac{44.5}{x^2} = \text{ anything ?}$
 
skeeter said:
what you have posted is an expression in two variables, not an equation

does $\dfrac{0.149}{(18 - 0.1x - 0.05n)^2} - \dfrac{44.5}{x^2} = \text{ anything ?}$
I'm so so so sorry for the mistake, the above expression is equal to 0.
 
$\dfrac{0.149}{(18 - 0.1x - 0.05n)^2} - \dfrac{44.5}{x^2} = 0$

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.
 
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