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October 19th, 2016, 09:29 PM   #1
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Maths Irrational

Express 14-8√60÷6√20-10√12 in the form a√b+6√3; where a & b are irrational numbers.
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October 19th, 2016, 09:59 PM   #2
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Express 14-8√60÷6√20-10√12 in the form a√b+6√3; where a & b are irrational numbers.
$\dfrac{14-8\sqrt{60}}{6\sqrt{20}-10\sqrt{12}}$

$\dfrac{14-8\sqrt{60}}{6\sqrt{20}-10\sqrt{12}}\dfrac{6\sqrt{20}+10\sqrt{12}}{6\sqrt{ 20}+10\sqrt{12}} =$

$\dfrac{(14-8\sqrt{60})(6\sqrt{20}+10\sqrt{12})}{36\cdot 20 - 100\cdot 12}=$

$\dfrac{(14-16\sqrt{15})(12\sqrt{5}+20\sqrt{3})}{-480}=$

$\dfrac{-8}{480}(7-8\sqrt{15})(3\sqrt{5}+5\sqrt{3}) =$

$-\dfrac{1}{60}(21\sqrt{5}-24\sqrt{75}+35\sqrt{3}-40\sqrt{45})=$

$-\dfrac{1}{60}(21\sqrt{5}-120\sqrt{3}+35\sqrt{3}-120\sqrt{5}) =$

$-\dfrac{1}{60}(-99\sqrt{5}-85\sqrt{3}) = \dfrac{99}{60}\sqrt{5} + \dfrac{85}{60}\sqrt{3} = $

$\dfrac{33}{20}\sqrt{5}+\dfrac{17}{12}\sqrt{3}$
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October 20th, 2016, 06:59 AM   #3
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Thanks romsek, am most grateful!
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October 20th, 2016, 12:25 PM   #4
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Quote:
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Express 14-8√60÷6√20-10√12 in the form a√b+6√3; where a & b are irrational numbers.
Your problem is stated wrong. You were asked multiple times to use parentheses around your expressions and to show some work.

You were shown a similar model problem at least one time before that you could have started your work.


And, romsek, you shouldn't have given a complete solution only three hours later without the user showing any work.
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November 4th, 2016, 05:28 PM   #5
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... you shouldn't have given a complete solution only three hours later without the user showing any work.
I encourage all members to post freely, regardless of their personal philosophy. The sooner we learn to respect each other's differences is the sooner we all start to get along.

As the OP did not post back asking for help and/or clarification, I don't think it's unreasonable to assume that they understood what has been posted. Text books provide worked examples; why can't we?

MMBt, it is not your place to tell other users what they should and should not post. You are not an admin.
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Last edited by skipjack; November 5th, 2016 at 01:30 PM.
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November 4th, 2016, 05:58 PM   #6
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And, romsek, you shouldn't have given a complete solution only three hours later without the user showing any work.
What in the world lol. The amount of threads that have been answered without OP providing any working is large. And yet you randomly choose this one? Please, the trolling needs to stop. New users get so put off by this sort of thing.
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November 4th, 2016, 06:59 PM   #7
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And, romsek, you shouldn't have given a complete solution only three hours later without the user showing any work.
Back then I also liked to provide a full solution if I could. However, I didn't intend to help the posters. The posters I seek at that time were the ones who desperately looked for answers without showing any works and had some deadlines. I just wanted to test my math skills and their procrastination was the perfect sacrifice for it.
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November 5th, 2016, 01:07 AM   #8
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Getting back to the matter at hand...

Is there any a priori reason that says that pi should be irrational? Is there something in Mathematics that would cause it to be so?

-Dan
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November 5th, 2016, 05:20 AM   #9
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Well, all the Proofs that $\pi$ is irrational are purely deductive, aren't they?

It's an odd question. Perhaps I have misunderstood your usage of "a priori", although I did look it up to check the meaning.
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November 5th, 2016, 05:20 AM   #10
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Well, all the Proofs that $\pi$ is irrational are purely deductive, aren't they?

It's an odd question. Perhaps I have misunderstood your usage of "a priori", although I did look it up to check the meaning.
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