MHB Which side has a larger value?

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The discussion revolves around determining whether the sum of the square roots of 2, 5, and 11 is greater than or less than 7, without using the square root function on a calculator. Participants explore alternative methods, including algebraic manipulation and squaring both sides of the inequality to simplify the comparison. The calculations lead to the conclusion that the sum of the square roots is indeed less than 7. While some participants express the challenge of performing these calculations without a calculator, others emphasize that it is still possible with traditional methods. The thread highlights the balance between convenience and mathematical skill in solving such inequalities.
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Without using the square root button on a calculator,
determine which side has a larger value:\sqrt{2} \ + \sqrt{5} \ + \sqrt{11}\ versus \ \ 7
 
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Interesting. Can you use all other buttons on your calculator? If there is a solution without using the calculator then I would like to see that very much. Guess, check and adjust isn't very elegant.
 
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Jameson said:
Interesting.
Can you use all other buttons on your calculator? If there is a solution without using the calculator then I
would like to see that very much. Guess, check and adjust
isn't very elegant.

Only these keys** may be used:
-------------------------------

add

subtract

multiply

divide

parentheses

memory store

memory recall

equals/enter button

- - - - - - - - - - - - - - - - - - - - - - - -

And, you may use paper and something
with which to write on the paper.<><><><><><><><><><><><><><><><><>**This also includes, as an example, that you
cannot use an exponentiation key, such as
y^x. And then that eliminates the possible uses
of x^(1/2) and/or x^(0.5).
<> <> <> My solution may be forthcoming in a 1/2 day to 2 days
from now, so I could give users a chance.
 
checkittwice said:
Without using the square root button on a calculator,
determine which side has a larger value:\sqrt{2} \ + \sqrt{5} \ + \sqrt{11}\ versus \ \ 7
In the inequality $\sqrt{2} + \sqrt{5} + \sqrt{11} \; \diamondsuit\; 7$, you have to decide whether the $\diamondsuit$ symbol should be < or >. Start by subtracting $\sqrt2$ from both sides: $\sqrt{5} + \sqrt{11} \ \diamondsuit\ 7 - \sqrt{2}.$ Now square both sides: $16 + 2\sqrt{55} \ \diamondsuit\ 51 - 14\sqrt2$.

Thus $2\sqrt{55} + 14\sqrt2 \ \diamondsuit\ 35$. Now square both sides again: $612 + 56\sqrt{110} \ \diamondsuit\ 1225$, and therefore $56\sqrt{110} \ \diamondsuit\ 613$.

So far, that has scarcely even needed a calculator. The last step is to square both sides again, and for that you do need the calculator, to get $344960 \ \diamondsuit\ 375769$, from which it is clear that $\diamondsuit$ has to be <.
 
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Opalg said:
So far, that has scarcely even needed a calculator. The last step is to square both sides again, and for that you do need the calculator, to get $344960 \ \diamondsuit\ 375769$, from which it is clear that $\diamondsuit$ has to be <.

It might be convenient to use a calculator, but surely one does not need to use a calculator!?

CB
 
CaptainBlack said:
It might be convenient to use a calculator, but surely one does not need to use a calculator!?

CB
I suppose that depends on whether one remembers (or was ever taught) how to do long multiplication. (Giggle)
 
Opalg said:
I suppose that depends on whether one remembers (or was ever taught) how to do long multiplication. (Giggle)

It is still taught (after a fashion) in UK junior schools (as of a few (<5) years ago) when my children were at that stage of their educations.

CB
 
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