Basic Algebra: Showing E = .5(k/a) Without The Conjugate

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SUMMARY

The equation E = .5(k/a)(e^2 - 1)/(1 - e^2) simplifies to E = .5(k/a) under specific algebraic manipulations. The discussion highlights that factoring -1 from the denominator allows for the cancellation of (e^2 - 1), leading to the simplified form. The conjugate method is deemed ineffective for this particular equation. Participants confirmed that the correct interpretation of the equation is essential for accurate simplification.

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Nusc
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Can someone explain to me why E = .5(k/a)(e^2 - 1)/(1 - e^2) = .5(k/a)

The conjugate won't work, how do I show this?
 
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Nusc said:
Can someone explain to me why E = .5(k/a)(e^2 - 1)/(1 - e^2) = .5(k/a)
The conjugate won't work, how do I show this?

You can factor -1 out of the denominator and then cancel (e^2-1), but that leaves you with E = .5(k/a)(e^2 - 1)/(1 - e^2) = -.5(k/a)

GM
 
Actually that was what I meant 0.5(k/a), thanks
 

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