How to Simplify an Algebraic Equation in One Step?

  • Thread starter Thread starter phospho
  • Start date Start date
  • Tags Tags
    Algebra
AI Thread Summary
To simplify the algebraic equation 2a(q-t) + at(q-t)(q+t) = 0 in one step, one can factor out a(q-t) from both terms. This results in a(q-t)(2 + t(q+t)) = 0. The key is recognizing that both terms share the common factor a(q-t). The remaining expression after factoring is 2 + t(q+t). Understanding this factoring technique is essential for simplifying similar algebraic equations efficiently. Mastering such steps can greatly enhance problem-solving skills in algebra.
phospho
Messages
250
Reaction score
0
vq154z.png


Could anyone explain how they made the jump from the highlighted parts? i.e how do you go from

2a(q-t) + at(q-t)(q+t) = 0 to
a(q-t)(2+t(q+t)) = 0

in one step
 
Physics news on Phys.org
If you pull out a(q-t) from both terms of the summation in the first equation, what do you have left?
 
thanks
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top