How is the quotient of two constants calculated in a given equation?

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Homework Statement


Part of a larger problem. I know that
[tex]F_{1}^2+2F_{1}F_{2}-F_{2}^2=0[/tex]
where [tex]F_{1}[/tex] and [tex]F_{2}[/tex] are x and y components of a force. Hence
[tex]\frac{F_{1}}{F_{2}}=1\pm\sqrt{2}[/tex]
I can't see how that step is done.

Homework Equations



The Attempt at a Solution


[tex]{F_{1}=\sqrt{F_{2}^2+2F_{1}F_{2}}[/tex]
[tex]F_{2}=\sqrt{F_{1}^2-2F_{1}F_{2}}[/tex]
Stuck here.
Thanks
 
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If you were asked to solve [tex]x^2+2x=0[/tex] for x, you wouldn't say [tex]x=\sqrt{-2x}[/tex] but would rather solve it using the quadratic formula. Do the same for F1 and F2, you can treat F1 as the variable x that you need to solve and F2 as a constant, and vice versa.
 
Mentallic said:
If you were asked to solve [tex]x^2+2x=0[/tex] for x, you wouldn't say [tex]x=\sqrt{-2x}[/tex] but would rather solve it using the quadratic formula. Do the same for F1 and F2, you can treat F1 as the variable x that you need to solve and F2 as a constant, and vice versa.

I did actually try that way also, with answer:
[tex]\frac{F_{1}}{F_{2}}=\frac{F_{2}\pm\sqrt{1+F_{2}^2}}{-F_{1}\pm\sqrt{1+F_{1}^2}}[/tex]
I can't see how to simplify that either. But since F1 and F2 are constants, isn't it meaningless to take their roots?
Thanks
 
fysiikka111 said:
I did actually try that way also, with answer:
[tex]\frac{F_{1}}{F_{2}}=\frac{F_{2}\pm\sqrt{1+F_{2}^2}}{-F_{1}\pm\sqrt{1+F_{1}^2}}[/tex]
Note quite, if we have [tex]x^2+2xy-y^2=0[/tex] and solve for x, we get [tex]x=\frac{-2y\pm\sqrt{4y^2+4y^2}}{2}=y\left(-1\pm\sqrt{2}\right)[/tex]

fysiikka111 said:
I can't see how to simplify that either. But since F1 and F2 are constants, isn't it meaningless to take their roots?
Thanks

What do you mean take their roots? As in use the quadratic formula to solve them? They can be thought of as variables even though they are actually constant. We are able to interchange between F1 being a variable and F2 the constant and vice versa.
 
fysiikka111 said:

Homework Statement


Part of a larger problem. I know that
[tex]F_{1}^2+2F_{1}F_{2}-F_{2}^2=0[/tex]
where [tex]F_{1}[/tex] and [tex]F_{2}[/tex] are x and y components of a force. Hence
[tex]\frac{F_{1}}{F_{2}}=1\pm\sqrt{2}[/tex]
I can't see how that step is done.

Homework Equations



The Attempt at a Solution


[tex]{F_{1}=\sqrt{F_{2}^2+2F_{1}F_{2}}[/tex]
[tex]F_{2}=\sqrt{F_{1}^2-2F_{1}F_{2}}[/tex]
Stuck here.
Thanks

To add to what Mentallic said, when you solve for a variable, you end up with an equation in which the variable appears by itself on one side and does not appear on the other side. You have not solved for F1 if it appears on both sides of your final equation. Same for F2.

For example, if x2 + 2x - 1 = 0, you have not solved for x if you rewrite this equation as
[tex]x = \pm \sqrt{1 - 2x}[/tex]
 
I would have done it this way:
Complete the square
[tex]F_1^2+ 2F_1F_2- F_2^2= F_1^2+ 2F_1F_2+ F_2^2- 2F_2^2[/tex]
[tex]= (F_1+ F_2)^2- 2F_2^2[/tex]
a "difference of squares" which can be written a "product of sum and difference" so
[tex](F_1+ F_2)^2- 2F_2^2= (F_1+ F_2- \sqrt{2}F_2)(F_1+ F_2+ \sqrt{2)F_2)= 0[/tex]
so we must have either
[tex]F_1+ F_2- \sqrt{2}F_2= F_1+ (1-\sqrt{2})F_2= 0[/tex]
so that [itex]F_1= (1-\sqrt{2})F_2[/itex] and so
[tex]\frac{F_1}{F_2}= 1- \sqrt{2}[/tex]

or
[tex]F_1+ F_2+ \sqrt{2}F_2= F_1+ (1+\sqrt{2}) F_2= 0[/tex]
so that [itex]F_1= (1+ \sqrt{2})F_2[/itex] and so
[tex]\frac{F_1}{F_2}= 1+ \sqrt{2}[/tex]
 
HallsofIvy said:
I would have done it this way:
Complete the square
[tex]F_1^2+ 2F_1F_2- F_2^2= F_1^2+ 2F_1F_2+ F_2^2- 2F_2^2[/tex]
[tex]= (F_1+ F_2)^2- 2F_2^2[/tex]
a "difference of squares" which can be written a "product of sum and difference" so
[tex](F_1+ F_2)^2- 2F_2^2= (F_1+ F_2- \sqrt{2}F_2)(F_1+ F_2+ \sqrt{2)F_2)= 0[/tex]
so we must have either
[tex]F_1+ F_2- \sqrt{2}F_2= F_1+ (1-\sqrt{2})F_2= 0[/tex]
so that [itex]F_1= (1-\sqrt{2})F_2[/itex] and so
[tex]\frac{F_1}{F_2}= 1- \sqrt{2}[/tex]

or
[tex]F_1+ F_2+ \sqrt{2}F_2= F_1+ (1+\sqrt{2}) F_2= 0[/tex]
so that [itex]F_1= (1+ \sqrt{2})F_2[/itex] and so
[tex]\frac{F_1}{F_2}= 1+ \sqrt{2}[/tex]

Ahh much more elegant :wink:
Just a little query - your math says [tex]F_1+(1-\sqrt{2})F_2=0[/tex] so solving from this we should get [tex]\frac{F_1}{F_2}=-(1-\sqrt{2})[/tex] and also [tex]F_1+(1+\sqrt{2})F_2=0[/tex] thus [tex]\frac{F_1}{F_2}=-(1+\sqrt{2})[/tex]

I'm probably missing something blindingly obvious here...