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

  • Thread starter thomasrules
  • Start date
In summary, Daniel says that when solving for x in the equation 0.012M/(d-x)^2=M/x^2, you should consider the modulus when doing so.
  • #1
thomasrules
243
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
 
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  • #2
show your work
 
  • #3
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
 
  • #4
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.
 
  • #5
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.
 
  • #6
does d in this case stand for difference like in calculus or is it just any other variable?
 
  • #7
By,the looks of it,it's a dimensionful number...

Daniel.

PS.Got it,it's the mean distance Earth-Moon and M is the Earth's mass... :tongue2: :approve: And it's in Km.
 
  • #8
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??
 
  • #9
I'm sorry,but it doesn't work that way.You need to consider the modulus when extracting sqrt,which would complicate the problem...

Daniel.
 
  • #10
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
 
  • #11
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... :grumpy:


Daniel.
 
  • #12
if its correct please tell me. :smile:
 
  • #13
I thought i put in a very obvious way... :tongue2: It's wrong.By inappropriately extracting square root,u lost one (possibly viable) solution.

Daniel.
 

What is the purpose of solving for x in the equation 0.012M/(d-x)^2=M/x^2?

The purpose of solving for x in this equation is to find the value of x that will make the equation true. This value of x is known as the solution or root of the equation.

What is the significance of the numbers 0.012, M, and d in the equation?

The number 0.012 is the coefficient of the fraction in the numerator and represents a constant value in the equation. M and d are variables that represent unknown values in the equation.

Why are there two different values of x in the equation?

The equation contains two different values of x because the fraction in the numerator and denominator are being squared. This allows for the possibility of two different solutions for x.

How can I solve this equation for x?

To solve for x, you can use algebraic techniques such as cross-multiplication, factoring, or the quadratic formula. You can also use a graphing calculator to find the intersection point of the two sides of the equation.

What is the practical application of this equation?

This equation can be used in various scientific fields, such as chemistry and physics, to calculate unknown values related to concentration, distance, and mass. It can also be used in engineering and economics to solve optimization problems.

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