Finding the Root of a Quadratic Equation with Factoring Difficulty

In summary, the problem involves finding the net gravitational force exerted by two objects with masses of 200kg and 500kg on a 50kg object placed at a specific position between them. The equations used are based on the Law of Universal Gravitation and involve factoring and solving for unknown variables. The solution for part b) is between the two objects and 0.245m from the 500kg object. However, the individual solving the problem is still having difficulties and is seeking further assistance.
  • #1
AznBoi
471
0

Homework Statement


Objects with masses of 200kg and 500kg are separated by 0.4m a)Find the net gravitational force exerted by these objects on a 50kg object placed midway between them. b) At what position (other than infinitely remote ones) can the 50 kg object be placed so as to experience a net force of zero?

I have solved a), everything below is for part b)

I'm solving for a problem right now and I'm having difficulties in factoring. =P

Trying out Latex, it may take a while.

Homework Equations


[tex]0=\frac{Gm_{1}m_{3}}{-r_{1}{}^2} + \frac{Gm_{2}m_{3}}{r_{2}{}^2}[/tex]


The Attempt at a Solution


[tex]0=\frac{Gm_{1}m_{3}}{-r_{1}{}^2} + \frac{Gm_{2}m_{3}}{r_{2}{}^2}[/tex]

[tex]0=Gm_{3}(\frac{m_1}{-r_{1}{}^2} + \frac{m_2}{r_{2}{}^2})[/tex]

[tex]0=\frac{m_1}{-r_{1}{}^2} + \frac{m_2}{r_{2}{}^2}[/tex]

[tex]0=m_{1}r_{2}{}^2 - m_{2}r_{1}{}^2[/tex]

[tex]0=(200{}kg)r_{2}{}^2 - (500{}kg)r_{1}{}^2[/tex]

Since there are two unknown variables, you need a second equation:

[tex]0.4m=r_{1} + r_{2}[/tex]

~~~~~~~~~~~~~~~~~~~~~~~~~~~


I just don't know how to factor:
[tex]0=(200{}kg)r_{2}{}^2 - (500{}kg)r_{1}{}^2[/tex] to make it in [tex](r_{1}+r_{2})(r_{1} - r_{2})[/tex] form. Can someone show me the simple factoring that I've forgotten? xP
 
Last edited:
Physics news on Phys.org
  • #2
By the way, are my previous factoring steps correct? Thanks!
 
  • #3
AznBoi said:

Homework Statement


Objects with masses of 200kg and 500kg are separated by 0.4m a)Find the net gravitational force exerted by these objects on a 50kg object placed midway between them. b) At what position (other than infinitely remote ones) can the 50 kg object be placed so as to experience a net force of zero?

I have solved a), everything below is for part b)

I'm solving for a problem right now and I'm having difficulties in factoring. =P

Trying out Latex, it may take a while.

Homework Equations


[tex]0=\frac{Gm_{1}m_{3}}{-r_{1}{}^2} + \frac{Gm_{2}m_{3}}{r_{2}{}^2}[/tex]

The Attempt at a Solution


[tex]0=\frac{Gm_{1}m_{3}}{-r_{1}{}^2} + \frac{Gm_{2}m_{3}}{r_{2}{}^2}[/tex]

[tex]0=Gm_{3}(\frac{m_1}{-r_{1}{}^2} + \frac{m_2}{r_{2}{}^2})[/tex]

[tex]0=\frac{m_1}{-r_{1}{}^2} + \frac{m_2}{r_{2}{}^2}[/tex]

[tex]0=m_{1}r_{2}{}^2 - m_{2}r_{1}{}^2[/tex]

[tex]0=(200{}kg)r_{2}{}^2 - (500{}kg)r_{1}{}^2[/tex]

Since there are two unknown variables, you need a second equation:

[tex]0.4m=r_{1} + r_{2}[/tex]

~~~~~~~~~~~~~~~~~~~~~~~~~~~


I just don't know how to factor:
[tex]0=(200{}kg)r_{2}{}^2 - (500{}kg)r_{1}{}^2[/tex] to make it in [tex](r_{1}+r_{2})(r_{1} - r_{2})[/tex] form. Can someone show me the simple factoring that I've forgotten? xP

I don't think you need to factor it, just do the following:

(hold on...)
 
Last edited:
  • #4
Well I think it would be easier if you factored it and then added/subtracted the two equations to find one unknown variable right?

Do I just substitute the equation you found with r_1 into the other equation?
 
  • #5
This is what I do when I get these kinds of equations: I plug them into my graphing calculator, graph them, and see where y = 0 to find the roots. Call me lazy, but it works :P So you can plug in the equation

[tex]y=(200)(0.4 - x)^2 - (500)x^2[/tex]

and look for where y = 0.
 
Last edited:
  • #6
gabee said:
This is what I do when I get these kinds of equations: I plug them into my graphing calculator, graph them, and see where y = 0 to find the roots. Call me lazy, but it works :P

Good idea! xD I've never tried that before with any physics equations. Hmm, mabye that's what I should do when I get stuck with these kinds of equations. Btw, they allow graphing calculators on the physics subject test lol! hehe. Thanks for your advice and help!

While I try that, if anyone would be willing to factor:

[tex]0=(200{}kg)r_{2}{}^2 - (500{}kg)r_{1}{}^2[/tex]

into the (x+y)(x-y) form, it would be greatly appreciated! :smile:
 
  • #7
You know how to factor x² - y², so all you have to do is to rewrite what you actually have in that form.
 
  • #8
Hurkyl said:
You know how to factor x² - y², so all you have to do is to rewrite what you actually have in that form.

How do you factor 200x² - 500y²?? Seriously, I'm uncertain of how to factor that lol.
 
  • #9
Ahh, now I've got it. The only slightly difficult part comes with the coefficients. Take the square roots of them, so that when they are multiplied together you get them back.

[tex]
0 = 100(2r_1^2 - 5r_2^2)
[/tex]

[tex]
0 = (\sqrt{2}r_1 + \sqrt{5}r_2)(\sqrt{2}r_1 - \sqrt{5}r_2)
[/tex]

[tex]
0 = (\sqrt{2}(0.4 - r_2) + \sqrt{5}r_2)(\sqrt{2}(0.4 - r_2) - \sqrt{5}r_2)
[/tex]

[tex]
0 = (0.4 \sqrt{2} + r_2 (\sqrt{5} - \sqrt{2}))(0.4 \sqrt{2} - r_2 (\sqrt{2} + \sqrt{5}))
[/tex]

Now you can find the roots of this equation.
 
Last edited:
  • #10
[tex]x^{2}-y^{2}=(x+y)(x-y)[/tex]

Don't take my word for it, expand it so that you understand.
 
  • #11
Just to demonstrate as maybe the coefficients are screwing you up...

[tex]200x^{2}-500x^{2}[/tex]
[tex]=100(2x^2-5y^2)[/tex] (might help if there is a 0 on the other hand, makes factoring simpler, although of course unnecessary)
[tex]=100(\sqrt{2}x+\sqrt{5}y)(\sqrt{2}x-\sqrt{5}y)[/tex]

in this case the "[tex]x^{2}[/tex]" in [tex]x^{2}-y^{2}=(x+y)(x-y)[/tex] is [tex]2x^{2}[/tex] and because
[tex]\sqrt{xy}=\sqrt{x}*\sqrt{y}[/tex]


[tex]\sqrt{2x^{2}}=\sqrt{2}*\sqrt{x^{2}}=\sqrt{2}x[/tex]
 
Last edited:
  • #12
This is a good skill to get used to. If you have an expression like

200 r12 - 500 r22

and you think you want to apply an identity like

x2 - y2 = (x + y) (x - y)

then you don't need to be creative: you can simply solve for what you need to do. Set

x2 = 200 r12
y2 = 500 r22

and then if you can solve for x and y, that shows you how you can apply your identity to your expression.
 
  • #13
I see now.. Yeah the coefficients were screwing me up =P Thanks for all your guys' help! :smile:
 
  • #14
I still don't get the answer that is posted in the back of the book with the factored equations and the second one. I don't know what I'm doing wrong. The two answers I got for r_2 is -0.6883m and -.55m... The answer for b) is Between the two objects and 0.245m from the 500kg object.
 
  • #15
:frown:
AznBoi said:
I still don't get the answer that is posted in the back of the book with the factored equations and the second one. I don't know what I'm doing wrong. The two answers I got for r_2 is -0.6883m and -.55m... The answer for b) is Between the two objects and 0.245m from the 500kg object.

I still am unable to solve b) At what position (other than infinitely remote ones) can the 50 kg object be placed so as to experience a net force of zero?

I have no idea what I'm doing wrong. I have came up with 2 equations that should relate to what is being asked. I have factored one of them so I can plug the other into find one of the lengths.. My answer is not correct however.

The answer for b) is Between the two objects and 0.245m from the 500kg object. he two answers I got for r_2 is -0.6883m and -.55m... :frown: Help Please? Thanks.
 
  • #16
gabee said:
Ahh, now I've got it. The only slightly difficult part comes with the coefficients. Take the square roots of them, so that when they are multiplied together you get them back.

[tex]
0 = 100(2r_1^2 - 5r_2^2)
[/tex]

[tex]
0 = (\sqrt{2}r_1 + \sqrt{5}r_2)(\sqrt{2}r_1 - \sqrt{5}r_2)
[/tex]

[tex]
0 = (\sqrt{2}(0.4 - r_2) + \sqrt{5}r_2)(\sqrt{2}(0.4 - r_2) - \sqrt{5}r_2)
[/tex]

[tex]
0 = (0.4 \sqrt{2} + r_2 (\sqrt{5} - \sqrt{2}))(0.4 \sqrt{2} - r_2 (\sqrt{2} + \sqrt{5}))
[/tex]

Now you can find the roots of this equation.

Check your arithmetic. The root of the second factor is positive.
 

1. What is factoring and why is it difficult?

Factoring is the process of finding the factors of a given number, which are the numbers that can be multiplied together to get that number. It is difficult because it involves finding all possible combinations of numbers that can multiply to the given number, which can be time-consuming and complex for large numbers.

2. What are some common strategies for factoring difficulties?

Some common strategies for factoring difficulties include using the greatest common factor, factoring by grouping, and using the difference of squares or cubes formulas. It may also be helpful to break down large numbers into smaller factors or to use trial and error.

3. How can factoring be used in real-life applications?

Factoring is used in various real-life applications, such as cryptography, where large numbers are factored to determine the security of encryption systems. It is also used in solving polynomial equations, finding the prime factorization of numbers, and simplifying algebraic expressions.

4. What are some common mistakes made when factoring?

Some common mistakes made when factoring include forgetting to check for common factors, making incorrect calculations or assumptions, and not considering all possible factors. It is also important to check the final answer to ensure that all factors have been accounted for.

5. Are there any techniques to make factoring easier?

Yes, there are some techniques that can make factoring easier. These include using a systematic approach, such as creating a table of factors, using patterns or shortcuts for certain types of numbers, and practicing regularly to improve speed and accuracy. It may also be helpful to break down larger numbers into smaller, more manageable factors.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
449
  • Introductory Physics Homework Help
Replies
8
Views
505
  • Introductory Physics Homework Help
Replies
19
Views
2K
  • Advanced Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
832
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
814
  • Introductory Physics Homework Help
Replies
15
Views
1K
Back
Top