How to i manipulate this fraction

  • Thread starter Thread starter seang
  • Start date Start date
  • Tags Tags
    Fraction
AI Thread Summary
To manipulate the fraction \(\frac{100K}{x^2 + (25 + \alpha)x + (\alpha 25 + 100K)}\) into the form \(\frac{K}{x^2 + \beta x + K}\), a scaling factor can be applied to either the numerator or denominator, ensuring both K values match. The discussion emphasizes that adjusting the denominator affects the x² term, which is undesirable. A proposed approach involves using a scaling factor of the form \(\frac{1}{C}\) to maintain equivalence while manipulating the terms. If scaling outside the fraction is not allowed, alternative methods must be explored to achieve the desired form. The conversation highlights the importance of maintaining the relationship between the numerator and denominator throughout the manipulation process.
seang
Messages
184
Reaction score
0

Homework Statement


I need to make this:

\frac{100K}{x^2 + (25 + \alpha)x + (\alpha 25 + 100K)}

look like this:

\frac{K}{x^2 + \beta x + (K)}

Homework Equations



Beta and K can be any combination of variables, as long as the top K and bottom K are equal. A scaling factor is allowed to come outside of the fraction if necessary.

The Attempt at a Solution

 
Physics news on Phys.org
The "scaling factor" means that you can multiply either numerator or denominator to make them the same. Multiplying the denominator by something would change the "x2" term and, apparently, you don't want that. That means you must have something of the form
\frac{1}{C}\frac{100CK}{x^2+ (25+\alpha)x+ (25\alpha+ 100K)}[/itex]<br /> It&#039;s pretty obvious what \beta must be! Can you choose C and the <b>new</b> K so that the numerator and constant term in the denominator are the same? (You might want to initially call the new K something like K&#039; to avoid confusion.)
 
alright, thank you.

But now, for the sake of argument, let's say I wasn't allowed to have a scaling factor outside of the fraction--what could I do then?
 
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...

Similar threads

Back
Top