Simplifying (2/6 -root3)^2 - (2/6+root3)^2

  • Thread starter Thread starter binomialgurl
  • Start date Start date
  • Tags Tags
    Simplifying
AI Thread Summary
The discussion centers on simplifying the expression (2/6 - √3)² - (2/6 + √3)². Participants suggest using the difference of squares formula, a² - b² = (a + b)(a - b), to simplify the problem effectively. There is a consensus that expanding the squared terms should be avoided initially, and instead, focus on rationalizing the denominator. The importance of adjusting both the numerator and denominator during simplification is emphasized. Overall, the goal is to evaluate the expression correctly while maintaining clarity in the steps taken.
binomialgurl
Messages
3
Reaction score
0
im having trouble trying to figure out a problem. can someone help me.

(2/6 -root3)^2 - (2/6+root3)^2

if that makes sense to anyone help.

I think that the denominator of both has to be rationalised but do i expand the squared brackets first or later or what?

any ideas welcome thanks
 
Physics news on Phys.org
I think you can use the difference of 2 squares a^2 - b^2 = ( a+b) (a-b). good luck!
 
Do you mean

\left(\frac{2}{6-\sqrt{3}}\right)^2 - \left(\frac{2}{6+\sqrt{3}}\right)^2

What are you supposed to do with this? Simplify?
 
Don't expand first. Use a^2 - b^2 = (a+b) (a-b) as tram said, at denominator, remember to do two times due to the square. This is to make denominator a whole number. Account for the multiplication for the numerator too.
 
I am supposed to evaluate it.
And thanks for all the advice by the way.
 
Last edited:
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