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 October 19th, 2016, 10:29 AM #1 Newbie   Joined: Oct 2016 From: Minna Posts: 8 Thanks: 0 Surds Express 14-8√60÷6√20-10√12 in the form a√b+6√3; where a & b are irrational numbers.
 November 4th, 2016, 04:54 AM #2 Global Moderator   Joined: Dec 2006 Posts: 18,956 Thanks: 1602 This question looks a bit strange. Where did it come from?
November 4th, 2016, 10:14 AM   #3
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Quote:
 Originally Posted by Hezekiah Express 14-8√60÷6√20-10√12 in the form a√b+6√3; where a & b are irrational numbers.
The obvious way to attack this problem, which looks VERY strange as skipjack said, is to set up an equation. What exactly IS the problem anyway? Please be exact.

$a\sqrt{b} + 6\sqrt{3} = \dfrac{14 - 8\sqrt{60}}{6\sqrt{20} - 10\sqrt{12}}\ or$

$a\sqrt{b} + 6\sqrt{3} = 14 - \dfrac{8\sqrt{60}}{6\sqrt{20}} - 10\sqrt{12}?$

November 4th, 2016, 10:31 AM   #4
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Quote:
 Originally Posted by Hezekiah Express 14-8√60÷6√20-10√12 in the form a√b+6√3; where a & b are irrational numbers.

Maths Irrational

 November 4th, 2016, 01:36 PM #5 Global Moderator   Joined: Dec 2006 Posts: 18,956 Thanks: 1602 The same question was indeed posted, and an answer was given that assumed missing parentheses as per JeffM1's first expression. Although the desired form of a√b+6√3 wasn't achieved, Hezekiah seemed pleased with the answer given, so we never found out the fully correct version of the question.
November 4th, 2016, 01:51 PM   #6
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Quote:
 Originally Posted by skipjack The same question was indeed posted, and an answer was given that assumed missing parentheses as per JeffM1's first expression. Although the desired form of a√b+6√3 wasn't achieved, Hezekiah seemed pleased with the answer given, so we never found out the fully correct version of the question.
It was also posted (with full solution) at MHF.

-Dan

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