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June 19th, 2014, 10:02 PM  #1 
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Which side is greater? my challenge #1
Determine which side has the larger value: $\displaystyle \sqrt{2 \ } \ + \ \sqrt{7 \ } \ + \ \sqrt{287 \ } \ \ \ versus \ \ \ 21$ You are not allowed to do it by estimating roots. You may square each side when appropriate. You can subtract or add radicals and/or nonradicals from each side as the case may be. When you are done with the process, you should have only an integer on each side of the word "versus" so you can answer the question. 
June 20th, 2014, 03:26 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,464 Thanks: 2038 
21 = 1.414 + 2.645 + 16.941 = √1.999396 + √6.996025 + √286.997481 < √2 + √7 + √287

June 20th, 2014, 07:08 AM  #3  
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Quote:
that I won't allow it to be done by estimating roots. This includes the estimates of the roots. My solution does not use any decimals. The problem is still open. Here is a simpler example of how it works: (√3 + √5) versus 4 (√3 + √5)^2 versus (4)^2 3 + 2√(15) + 5 versus 16 8 + 2√(15) versus 16 2√(15) versus 8 √(15) versus 4 [√(15)]^2 versus (4)^2 15 < 16 So √3 + √5 < 4. Or, to answer the question, the right side is greater. Last edited by Math Message Board tutor; June 20th, 2014 at 07:35 AM.  
June 20th, 2014, 07:20 AM  #4 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
The minimal polynomial of $\sqrt2+\sqrt7+\sqrt{287}$ is $P(x)=x^8  1184x^6 + 504920x^4  91695104x^2 + 5964163984$ so presumably we can compare $P(\sqrt2+\sqrt7+\sqrt{287})=0$ to $P(21)=258263$ to conclude that $\sqrt2+\sqrt7+\sqrt{287}>21.$ With more patience you could do this in the stepbystep way you ask for, I think. 
June 20th, 2014, 07:55 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,634 Thanks: 2620 Math Focus: Mainly analysis and algebra 
\begin{align*} \sqrt{2} + \sqrt{7} + \sqrt{287} \qquad &\text{versus} \qquad 21 \\ 2 + 2\sqrt{14} + 2\sqrt{574} + 7 + 2\sqrt{2009} + 287 \qquad &\text{versus} \qquad 441 \\ 2 + 2\sqrt{14} + 2\sqrt{574} + 7 + 2\sqrt{2009} + 287 &= 2 + \sqrt{56} + \sqrt{2296} + 7 + \sqrt{8036} + 287 \\ &\gt 2+\sqrt{49}+\sqrt{2209}+7+\sqrt{7921}+287 = 504 \\ &\gt 504 \end{align*} 
June 20th, 2014, 08:19 AM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,634 Thanks: 2620 Math Focus: Mainly analysis and algebra 
Of course, I had to estimate the roots to get to 2209 and 7921. The only difference was that I didn't include any decimal places.
Last edited by skipjack; June 20th, 2014 at 11:58 PM. 
June 20th, 2014, 11:07 PM  #7  
Global Moderator Joined: Dec 2006 Posts: 20,464 Thanks: 2038 
That's incorrect, as $2+\sqrt{49}+\sqrt{2209}+7+\sqrt{7921}+287 = 2 + 7 + 47 + 7 + 89 + 287 = 439$, not $504$. Quote:
21000 = 1414 + 2645 + 16941 = √1999396 + √6996025 + √286997481 < 1000√2 + 1000√7 + 1000√287, which implies the desired inequality. That uses knowledge of some large perfect squares, but doesn't explicitly use estimation of square roots. The kind of approach you wanted is quite easy, but eventually involves large squares. One compares (√7 + √287)² to (21  √2)², i.e., 7(42 + 2√41) to 443  42√2. It suffices now to compare 14√41 to 149  42√2, and squaring again means it suffices to compare 8036 to 25729  12516√2. After rearranging and squaring again, the comparison becomes 2(12516²) versus 17693², i.e., 313300512 versus 313042249, and the lefthand side is greater, as required. Not quite. You need to know a little more about the polynomial's zeros to justify that argument. Last edited by skipjack; June 21st, 2014 at 12:14 AM.  
June 22nd, 2014, 08:01 AM  #8 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Agreed. I just didn't feel like working through the details that there were no zeros between the two points. You can do it by studying the derivative (and its derivative?).


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