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

[fysiikka111]

Homework Statement

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

Homework Equations

The Attempt at a Solution

{F_{1}=\sqrt{F_{2}^2+2F_{1}F_{2}}
F_{2}=\sqrt{F_{1}^2-2F_{1}F_{2}}
Stuck here.
Thanks

 

[Mentallic]

If you were asked to solve x^2+2x=0 for x, you wouldn't say x=\sqrt{-2x} 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.

 

[fysiikka111]

If you were asked to solve x^2+2x=0 for x, you wouldn't say x=\sqrt{-2x} 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:
\frac{F_{1}}{F_{2}}=\frac{F_{2}\pm\sqrt{1+F_{2}^2}}{-F_{1}\pm\sqrt{1+F_{1}^2}}
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

 

[Mentallic]

I did actually try that way also, with answer:
\frac{F_{1}}{F_{2}}=\frac{F_{2}\pm\sqrt{1+F_{2}^2}}{-F_{1}\pm\sqrt{1+F_{1}^2}}

Note quite, if we have x^2+2xy-y^2=0 and solve for x, we get x=\frac{-2y\pm\sqrt{4y^2+4y^2}}{2}=y\left(-1\pm\sqrt{2}\right)

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]

Thanks!

 

[Mark44]

Homework Statement

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

Homework Equations

The Attempt at a Solution

{F_{1}=\sqrt{F_{2}^2+2F_{1}F_{2}}
F_{2}=\sqrt{F_{1}^2-2F_{1}F_{2}}
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
x = \pm \sqrt{1 - 2x}

 

[HallsofIvy]

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

or
F_1+ F_2+ \sqrt{2}F_2= F_1+ (1+\sqrt{2}) F_2= 0
so that F_1= (1+ \sqrt{2})F_2 and so
\frac{F_1}{F_2}= 1+ \sqrt{2}

 

[Mentallic]

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

or
F_1+ F_2+ \sqrt{2}F_2= F_1+ (1+\sqrt{2}) F_2= 0
so that F_1= (1+ \sqrt{2})F_2 and so
\frac{F_1}{F_2}= 1+ \sqrt{2}

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

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