Algebraic conversion of LHS to RHS equation

  • Thread starter Thread starter physicsernaw
  • Start date Start date
AI Thread Summary
The user is struggling to convert the left-hand side (LHS) of a given equation to match the right-hand side (RHS). They attempted various methods, including manipulating denominators and finding common factors, but reached a dead end. A user provided a hint about factoring using the difference of squares, which helped the original poster understand the solution. The discussion highlights the importance of algebraic manipulation techniques in solving equations. The original poster expressed gratitude for the assistance received.
physicsernaw
Messages
41
Reaction score
0

Homework Statement



I'm having trouble converting the LHS to the RHS of this equation:

http://www.wolframalpha.com/input/?i=%281%2F%28sqrt%285+%2B+h%29+%29-+1%2F%28sqrt%285%29%29%29%2Fh+%3D+-1%2F%28%28sqrt%285%29*sqrt%285%2Bh%29%29*%28sqrt%285%2Bh%29+%2B+sqrt%285%29%29%29

Homework Equations



N/A

The Attempt at a Solution



I've tried a bunch of different things. One thing I tried was bringing the h in the denominator to the denominators of the numerators (i.e. making it a factor of the numerators' denominators). With common denominators and factoring out h I end up getting:

[urlhttp://www.wolframalpha.com/input/?i=%28sqrt%285%29+-+sqrt%285%2Bh%29%29%2F%28h*%28sqrt%285%2Bh%29*sqrt%285%29%29%29[/url]

...And I'm stuck there. I would appreciate if someone could just show me step-by-step the solution (working the LHS to meet the RHS please).
 
Last edited:
Physics news on Phys.org
What is the problem?

ehild
 
(a-b)(a+b) = a^2-b^2 and a = \sqrt{a} \sqrt{a}.
 
I got it, thanks all.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top