Rational Completeing the Square

  • Thread starter Thread starter ISX
  • Start date Start date
  • Tags Tags
    Rational Square
AI Thread Summary
The discussion revolves around the process of completing the square in a quadratic equation. A participant struggles with matching the squares and realizes a mistake in squaring fractions, specifically confusing 5/6 with 5/12. Clarification is provided that the correct transformation involves using (b/2) squared rather than b squared. The participant acknowledges their misunderstanding and expresses gratitude for the assistance. This highlights the importance of careful fraction manipulation in algebraic transformations.
ISX
Messages
120
Reaction score
0

Homework Statement


51v52u.jpg

Homework Equations


9fxswl.jpg


The Attempt at a Solution


Alright so the solution is in the above pic, but I can't get anywhere close. You can see from the green circles that the "squares" aren't matching up. So I'm not sure if I can't multiply fractions anymore or what.
2ivzmoj.jpg
 
Physics news on Phys.org
ISX said:

Homework Statement


[ IMG]http://i49.tinypic.com/51v52u.jpg[/PLAIN]

Homework Equations


[ IMG]http://i50.tinypic.com/9fxswl.jpg[/PLAIN]

The Attempt at a Solution


Alright so the solution is in the above pic, but I can't get anywhere close. You can see from the green circles that the "squares" aren't matching up. So I'm not sure if I can't multiply fractions anymore or what.
[ IMG]http://i47.tinypic.com/2ivzmoj.jpg[/PLAIN]
Your line

\displaystyle x^2-\frac{5}{6}x+\frac{25}{144}=-\frac{1}{6}+\frac{25}{144}

is correct.

What your last line shows (correctly) is that \displaystyle \left(x-\frac{5}{6}\right)^2\ne x^2-\frac{5}{6}x+\frac{25}{144}\ .

When you square 5/6, you don't get 25/144 .

How did you get 25/144 in the first place? You squared 5/12 not 5/6 .
 

Attachments

  • 2ivzmoj.jpg
    2ivzmoj.jpg
    31.3 KB · Views: 378
you went from
x^2 +bx + (\frac{b}{2})^{2}
to
(x+b)^{2}

and not to

(x+\frac{b}{2})^{2}
 
Hellllllll I've been doing it wrong for centuries then. Let's try this again.

2exspdu.jpg


2128lz5.jpg


Ahhhh! Thanks for the help guys! I was always just dropping the last number off and went with the first 2. Guess I never caught on that fractions didnt work that way.
 
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