My Math Forum Evaluation of surds

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 October 18th, 2016, 08:38 AM #1 Newbie   Joined: Oct 2016 From: Minna Posts: 8 Thanks: 0 Evaluation of surds Evaluate (1+√2÷√5+√3)+(1-√2÷√5-√3)
October 18th, 2016, 10:52 AM   #2
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Quote:
 Evaluate (1+√2÷√5+√3)+(1-√2÷√5-√3)
if you mean ...

$\dfrac{1+\sqrt{2}}{\sqrt{5}+\sqrt{3}} + \dfrac{1-\sqrt{2}}{\sqrt{5}-\sqrt{3}}$

common denominator is $(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})$ ...

$\dfrac{(1+\sqrt{2})(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})} + \dfrac{(1-\sqrt{2})(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}$

can you take it from here?

 October 19th, 2016, 01:15 AM #3 Newbie   Joined: Oct 2016 From: Minna Posts: 8 Thanks: 0 pls don't be offended, expand further
 October 19th, 2016, 01:43 AM #4 Member   Joined: Sep 2016 From: India Posts: 88 Thanks: 30 $\dfrac{1+\sqrt{2}}{\sqrt{5}+\sqrt{3}}+\dfrac{1-\sqrt{2}}{\sqrt{5}-\sqrt{3}}$ $=\dfrac{(1+\sqrt{2})(\sqrt{5}-\sqrt{3})+(1-\sqrt{2})(\sqrt{5}+\sqrt{3})}{(\sqrt{5}+\sqrt{3})( \sqrt{5}-\sqrt{3})}$ $=\dfrac{(1+\sqrt{2})(\sqrt{5}-\sqrt{3})+(1-\sqrt{2})(\sqrt{5}+\sqrt{3})}{(\sqrt{5})^2-(\sqrt{3})^2}$ $=\dfrac{(1+\sqrt{2})(\sqrt{5}-\sqrt{3})+(1-\sqrt{2})(\sqrt{5}+\sqrt{3})}{5-3}$ $=\dfrac{\sqrt{5}-\sqrt{\not3}+\sqrt{\not10}-\sqrt{6}+\sqrt{5}+\sqrt{\not3}-\sqrt{\not10}-\sqrt{6}}{2}$ $=\dfrac{2\sqrt{5}-2\sqrt{6}}{2}$ $=\sqrt{5}-\sqrt{6}$ Last edited by deesuwalka; October 19th, 2016 at 01:53 AM.
 October 19th, 2016, 02:24 AM #5 Newbie   Joined: Oct 2016 From: Minna Posts: 8 Thanks: 0 Express 14-8√60÷6√20-10√12 in the form a√b+6√3; where a & b are irrational numbers.
October 19th, 2016, 03:46 AM   #6
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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.
Maybe deesuwalka will do this one for you, also ...

October 19th, 2016, 04:41 AM   #7
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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.
Could we at least get you to use parenthesis?
(14-8√60)÷(6√20-10√12)

The first step is to rationalize the denominator. The conjugate expression to 6√20-10√12 is 6√20+10√12. Multiply the top and bottom of the original expression and see where it leads.

-Dan

October 19th, 2016, 07:40 AM   #8
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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.
Please do not post a new problem in the same thread. Start a new thread for this problem.

You have a model from the previous problem, so show some work there.

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