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June 19th, 2014, 10:02 PM   #1
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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 non-radicals 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.
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June 20th, 2014, 03:26 AM   #2
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21 = 1.414 + 2.645 + 16.941 = √1.999396 + √6.996025 + √286.997481 < √2 + √7 + √287
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June 20th, 2014, 07:08 AM   #3
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Quote:
Originally Posted by skipjack View Post
21 = 1.414 + 2.645 + 16.941 = √1.999396 + √6.996025 + √286.997481 < √2 + √7 + √287
skipjack, your solution does not count (in my case), because I stated
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.
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June 20th, 2014, 07:20 AM   #4
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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 step-by-step way you ask for, I think.
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June 20th, 2014, 07:55 AM   #5
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\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*}
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June 20th, 2014, 08:19 AM   #6
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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.
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June 20th, 2014, 11:07 PM   #7
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That's incorrect, as $2+\sqrt{49}+\sqrt{2209}+7+\sqrt{7921}+287 = 2 + 7 + 47 + 7 + 89 + 287 = 439$, not $504$.

Quote:
Originally Posted by Math Message Board tutor View Post
This includes the estimates of the roots. My solution does not use any decimals.
You hadn't prohibited decimals and they can trivially be avoided by writing
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 left-hand side is greater, as required.

Quote:
Originally Posted by CRGreathouse View Post
. . . presumably we can compare $P(\sqrt2+\sqrt7+\sqrt{287})=0$ to $P(21)=-258263$ to conclude that $\sqrt2+\sqrt7+\sqrt{287}>21.$
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.
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June 22nd, 2014, 08:01 AM   #8
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Quote:
Originally Posted by skipjack View Post
Not quite. You need to know a little more about the polynomial's zeros to justify that argument.
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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