This post follows on from this post where the following principle was presented:

Fundamental Principle of Solving ‘Easy’ Equations

Identify what is difficult or troublesome about the equation and get rid of it. As long as you do the same thing to both numbers (the “Lhs” and the “Rhs”), the equation will be replaced by a simpler equation with the same solution.

There are a number of subtleties here: basically sometimes you get extra ‘solutions’ (that are not solutions at all), and sometimes you can lose solutions.

Let us write the squaring function, e.g. 6\mapsto 36, x\mapsto x^2 by f(x)=x^2 and the square-rooting function by x\mapsto \sqrt{x}. It appears that (x^2,\sqrt{x}) are an inverse pair but not quite exactly. While

\displaystyle 81\overset{\sqrt{x}}{\mapsto} 9\overset{x^2}{\mapsto}81 and

\displaystyle 7\overset{x^2}{\mapsto}49\overset{\sqrt{x}}{\mapsto}7,

check out O.K. note that

-4\overset{x^2}{\mapsto}+16\overset{\sqrt{x}}{\mapsto}=+4\neq -4,

does not bring us back to where we started.

This problem can be fixed by restricting the allowable inputs to x^2 to positive numbers only but for the moment it is better to just treat this as a subtlety, namely while (\sqrt{x})^2=x, \sqrt{x^2}=\pm x… in fact I recommend that we remember that with an x^2 there will generally be two solutions.

The other thing we look out for as much as possible is that we cannot divide by zero.

There are other issues around such as the fact that \sqrt{x}>0, so that the equation \sqrt{x}=-2 has no solutions (no, x=4 is not a solution! Check.). This equation has no solutions.

Often, in context, these subtleties are not problematic. For example, equations with no solutions rarely arise and quantities might be positive so that if we have \pm\sqrt{a}, only +\sqrt{a} need be considered (for example, a might be a length).

Highlighting these small subtleties is not therefore to scare you off but to allow you full mastery. Such as in the next example.

Example

Solve

\displaystyle \frac{V_B^2}{2g}-\frac{\left(\frac{V_BA_B}{A_A}\right)^2}{2g}=h_A-h_B.

Solution: Take out the common factor of 1/2g:

\displaystyle \Rightarrow \frac{1}{2g}\left(V_B^2-\frac{V_B^2A_B^2}{A_A^2}\right)=h_A-h_B

\displaystyle \Rightarrow V_B^2\left(1-\frac{A_B^2}{A_A^2}\right)=2g(h_A-h_B).

Assume that \displaystyle 1-\frac{A_B^2}{A_A^2} is non-zero (you can’t divide by zero).

Let us see what it means if this is zero:

1-\frac{A_A^2}{A_B^2}=0
\displaystyle\Rightarrow A_B^2-A_A^2=0
\displaystyle \underset{\text{difference of two squares}}{\Rightarrow} (A_B-A_A)(A_B+A_A)=0

\displaystyle \Rightarrow A_B-A_A=0 or A_B+A_A=0
\displaystyle\Rightarrow A_B=A_A\text{ or }A_B=-A_A.

Now perhaps from the context of the equation, both A_A and A_B are strictly positive (bigger than zero), so the second possibility can’t occur. The first possibility can happen: if A_B=A_A we have that the original equation reads:

\displaystyle \frac{V_B^2}{2g}-\frac{\left(\frac{V_BA_B}{A_A}\right)^2}{2g}=h_A-h_B
\displaystyle \Rightarrow \frac{V_B^2}{2g}-\frac{\left(\frac{V_BA_B}{A_B}\right)^2}{2g}=h_A-h_B
\displaystyle \Rightarrow \frac{V_B^2}{2g}-\frac{V_B^2}{2g}=h_A-h_B
\displaystyle \Rightarrow h_A-h_B=0
\displaystyle \Rightarrow h_A=h_B.

Therefore you can assume that A_A\neq A_B because if they are equal h_A=h_B and the equation can’t be solved for V_B.

So assuming A_B\neq A_A you do have:

\displaystyle V_B^2=\frac{2g(h_A-h_B)}{1-\frac{A_B^2}{A_A^2}}.

Now remember because of the square there are two solutions (i.e. x^2=a\Rightarrow x=\pm \sqrt{a} — you need plus and  minus). Therefore the correct implication is that:

\displaystyle V_B=\pm \sqrt{\frac{2g(h_A-h_B)}{1-\frac{A_B^2}{A_A^2}}}.

However, if you assume that V_B>0, then it cannot be the negative square root.

In fact, under these assumptions:

  • A_A,\,A_B,\,V_B>0
  • A_A\neq A_B,

we can tidy this up a little further. Multiply the number inside the square root by \displaystyle \frac{A_A^2}{A_A^2}=1:

\displaystyle V_B=\sqrt{\frac{2gA_A^2(h_A-h_B)}{A_A^2-A_B^2}}
\displaystyle =A_A\sqrt{\frac{2g(h_A-h_B)}{A_A^2-A_B^2}}.

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