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Please show how the second formula can be derived from the first one.

(1) enter image description here

(2)
enter image description here

  • 1
    Did you try to at least expand the top...? – 2017-01-31
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    Sorry guys. I really can't figure this out. I'm not great at maths. – 2017-01-31
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    What *do* you understand about these formulas? (This is an honest question.) – 2017-01-31

2 Answers 2

1

I trust you know how to simplify $(1)$ into the following:$$\frac{a+2b}{2e}-(\frac{c+d}{2e})^{2}-\frac{a}{2e}+\frac{c^2}{4e^2}+\frac{d^2}{4e^2}$$Then, you get:$$\frac{a+2b}{2e}-(\frac{c^2+2cd+d^2}{4e^2})-\frac{a}{2e}+\frac{c^2}{4e^2}+\frac{d^2}{4e^2}$$ Simplifying this by subtracting the similar terms gives:$$\frac{2b}{2e}-\frac{2cd}{4e^2}$$ And finally factoring the common factor gives:$$\frac{1}{e}(b-\frac{cd}{2e})$$

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    Damn i knew i was missing something obvious. Thank you very much. – 2017-01-31
1

To simplify the calculations, define the auxiliary variables: $$a′ =\cfrac{a}{2e}\,, \;\;b′=\cfrac{b}{2e}\,, \;\;c′=\cfrac{c}{2e}\,, \;\;d′=\cfrac{D}{2e}$$

Then the expression $(1)$ becomes:

$$ \require{cancel} \begin{align} \left(\cancel{a′} +2b′ −(c′ +d′ )^2 \right)−(\cancel{a′} −c′^{\,2} −d′^{\,2} ) & = 2 b' - \cancel{c′^{\,2}} -2c'd' −\bcancel{d′^{\,2}} + \cancel{c′^{\,2}} + \bcancel{d′^{\,2}} \\ & = 2 b' - 2 c'd' \\ & = \cancel{2} \cfrac{b}{\cancel{2}e} - \bcancel{2} \cfrac{c}{\bcancel{2}e}\cfrac{d}{2e} \\ & = \cfrac{1}{e}\left(b - \cfrac{cd}{2e}\right) \end{align} $$