Transforming Simultaneous Equations into Simplified Form for Easy Solving

[mconn86]

Hi, I have the following two equations of the form:

A^2 - B^2=hj(k^2) and 2AB=mjk

and I need to get them into the form

A=k.sqrt(hj/2).{sqrt((1)+(m/hk)^2)+1}^(1/2)

and

B=k.sqrt(hj/2).{sqrt((1)+(m/hk)^2)-1}^(1/2)

I've been trying in vain for a while now and its starting to annoy me coz I know it should be simple. Any ideas?

 

[CompuChip]

What about you just solve the second one for either A or B and plug it into the other one?
Didn't check it, but if they are really equivalent it should work.

 

[mconn86]

Yeah I've done that, that's not the problem really,its just getting it into the form I need.

 

[CompuChip]

I'm not sure I'll get exactly what you got, but I'll give it a try.

A^2 - B^2=hj(k^2) and 2AB=mjk

and I need to get them into the form

A=k.sqrt(hj/2).{sqrt((1)+(m/hk)^2)+1}^(1/2)

The second one gives B = mjk/2A. Plugging this into the first one gives
$$A^2 - \frac{ (m j k / 2)^2 }{ A^2 } = h j k^2$$
Calling $\alpha = A^2$ and multiplying through by $\alpha$, we get
$$\alpha^2 - h j k^2 \alpha - (m j k / 2)^2 = 0$$
The solutions are
$$\alpha = \frac12\left( h j k^2 \pm \sqrt{h^2 j^2 k^4 + j^2 k^2 m^2 \right)$$
so let's take the positive one. From now on I'll start taking things out of square roots by using $\sqrt{X Y^2} = Y \sqrt{X}$, so you need the assumption that they are positive. First rewrite alpha into
$$\alpha = \cdot \frac{1}{2} \left(h j k^2 + h j k^2 \sqrt{1 + m^2 / (h^2 k^2)} \right)$$
and then take out h j k^2:
$$\alpha = k^2 \cdot \frac{h j}{2} \left(1 + \sqrt{1 + m^2 / (h^2 k^2)} \right)$$
Solve $\alpha = A^2$ for A (pick the positive root again), take the k^2 outside the square root and use $\sqrt{a b} = \sqrt{a} \cdot \sqrt{b}$.