Quick way to simplify (12(sqrt(2) + 17)

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The discussion revolves around simplifying the expression 12(sqrt(2) + 17) and finding a quicker method to arrive at the precise answers provided in Mary L Boas' textbook. The original poster expresses frustration at not matching the textbook's answer, which is \frac{1}{32}(51\sqrt{2} - ln(1 + \sqrt{2}). A participant suggests recognizing the relationship between factors, specifically that (17 + 12√2)(17 - 12√2) equals 1, indicating a potential simplification method. The conversation highlights the importance of understanding factorization techniques to simplify expressions effectively. Ultimately, the thread emphasizes the value of exploring different approaches to problem-solving in mathematics.
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Homework Statement



This isn't really a problem at all. Have been working on some problems in Mary L Boas' textbook with some difficulty, but I have managed to solve the majority of the questions eventually. However, I've been stumped at how to arrive at the precise answers quoted in the book.

For instance, one of the answers is \frac{1}{32} (51\sqrt{2} - ln (1 + \sqrt{2}). But due to the method of arriving at the answer, my answer was \frac{1}{32}(51\sqrt{2}) - \frac{1}{128}(ln (17 + 2\sqrt{2}) which is equivalent (12 root 2 + 17 = (1 + rt 2)^4, after working backwards). But I'm wondering if there's actually a way to factorise or simplify my original (12 rt 2 + 17) quickly without working backwards.
 
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bigevil said:

Homework Statement



This isn't really a problem at all. Have been working on some problems in Mary L Boas' textbook with some difficulty, but I have managed to solve the majority of the questions eventually. However, I've been stumped at how to arrive at the precise answers quoted in the book.

For instance, one of the answers is \frac{1}{32} (51\sqrt{2} - ln (1 + \sqrt{2}). But due to the method of arriving at the answer, my answer was \frac{1}{32}(51\sqrt{2}) - \frac{1}{128}(ln (17 + 2\sqrt{2}) which is equivalent (12 root 2 + 17 = (1 + rt 2)^4, after working backwards). But I'm wondering if there's actually a way to factorise or simplify my original (12 rt 2 + 17) quickly without working backwards.

Hi bigevil! :smile:

(have a square-root: √ :wink:)

This is a bit of a hindsight way too …

you could have noticed that (17 + 12√2)(17 - 12√2) = 289 - 288 = 1,

so any factor was going to have the same property …

like 1 ± √2 and 3 ± 4√2 :smile:
 
Thanks tim! That's pretty clever...
 
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
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