Solving for X: 0.012M/(d-x)^2=M/x^2

  • Thread starter Thread starter thomasrules
  • Start date Start date
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
12 replies · 2K views
thomasrules
Messages
243
Reaction score
0
For some reason i can't solve for x:

0.012M/(d-x)^2=M/x^2

I have M value: (5.98)(10)^24

d is: (3.84)(10)^5
 
Physics news on Phys.org
i started with cross multiplying:

0.012Mx^2=M(d-x)^2

0.012Mx^2=M(d^2-2dx+x^2)

then i divided both sides by the left side and crossed out M.

From there I don't know what to do
 
After you divided both sides by M and then divided both sides by .012, you should have wound up with:

[tex]x^2=\frac{d^2-2dx+x^2}{.012}[/tex]

Split the right side up into three fractions:

[tex]x^2=\frac{d^2}{.012}-\frac{2dx}{.012}+\frac{x^2}{.012}[/tex]

Substitute in your value for d. That should leave you with something you can work with.
 
Actually, I wouldn't divide both sides by 0.012, just because I don't like fractions!

0.012x2= d2- 2dx+ x2 is the same as (subtract 0.012x2 from both sides)
0.988x2- 2dx+ d2= 0.

With d= 3.85 x 105, d2= 1.48225 x 1011 so your equation is
0.988x2- (7.7 x 105)x+ 1.48225 x 1011= 0.

You can solve that with the quadratic formula.
 
does d in this case stand for difference like in calculus or is it just any other variable?
 
couldnt you first divde M on both sides and then find the square root??

as in

[tex]\frac{0.012M}{(d-x)^2} = \frac{M}{x^2}[/tex]
which then gives

[tex]\frac{0.012}{(d-x)^2} = \frac{1}{x^2}[/tex]

and then find the square rooton both sides

[tex]\frac{\sqrt{0.012}}{d-x} = \frac{1}{x}[/tex]

then cross multiply and solve without need for a quadratic??
 
see i think it's this,
0.012M/(d-x)^2=M/x^2

cross multiply,

x^2/(d-x)^2=M/0.012M

x/(d-x)={1/0.012}^1/2 (that is sq root)

x/(d-x)=9.13

dividing the numerator & denominator by x

1/{(d/x) - 1}=9.13

{(d/x) - 1}=1/9.13
=0.11

(d/x)=1+.11
=1.11

since,d is: (3.84)(10)^5,we have

x=d/1.11
=(3.84)(10)^5/1.11
=345,945.95 units
 
jackal said:
see i think it's this,
0.012M/(d-x)^2=M/x^2

cross multiply,

x^2/(d-x)^2=M/0.012M

x/(d-x)={1/0.012}^1/2 (that is sq root)

THAT IS DEVIOUS...U should have a [tex]\pm[/tex] sign...


Daniel.
 
if its correct please tell me. :smile: