MHB Can You Factor Out (x-a)^(-3/2) in This Complex Equation?

  • Thread starter Thread starter mathdad
  • Start date Start date
  • Tags Tags
    Factoring
AI Thread Summary
The discussion focuses on factoring the complex equation involving terms with exponents of (x-a) and (x+a). The initial suggestion is to factor out (x-a)^(-3/2) as a starting point. However, another participant proposes factoring out (x-a)^(-1/2)(x+a)^(-3/2) instead. This alternative approach aims to simplify the equation further while maintaining the integrity of the original expression. The conversation emphasizes the importance of choosing the correct factor to facilitate easier manipulation of the equation.
mathdad
Messages
1,280
Reaction score
0
Factor.

I need help getting me started with this monster question.

[(x-a)^(-1/2)]/2 * (x+a)^(-1/2) - [(x+a)^(1/2)]/2 *(x-a)^(-3/2)

Do I factor out (x - a)^(-3/2) as step one?
 
Mathematics news on Phys.org
We are given to factor:

$$\frac{1}{2}(x-a)^{-\frac{1}{2}}(x+a)^{-\frac{1}{2}}-\frac{1}{2}(x-a)^{\frac{1}{2}}(x+a)^{-\frac{3}{2}}$$

As my first step, I would factor out:

$$\frac{1}{2}(x-a)^{-\frac{1}{2}}(x+a)^{-\frac{3}{2}}$$
 
That was my first choice to factor out.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...

Similar threads

Replies
5
Views
1K
Replies
2
Views
1K
Replies
1
Views
1K
Replies
3
Views
983
Replies
1
Views
1K
Replies
1
Views
1K
Replies
2
Views
1K
Replies
8
Views
1K
Back
Top