division of radicals.........

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January 23rd 2007, 02:04 AM
quebec567

division of radicals.........

Hi! I'm a canadian student in america.

Please help me on division of radicals =)

1.) 6√18 ÷ 12√40

2.) 8√19 ÷ 4√38

3.) 20√3 ÷ 5√ 3

4.) 42√ 6 ÷ 3√6

5.) -4√20 ÷ √2

6.) 1 / 2+√5

7.) 1 / 3-√11

8.) 3 / √3-1

9.) 5 / 2-√3

10.) 2 / 4-√14

thanks!
January 23rd 2007, 02:25 AM
AfterShock

Quote:

Originally Posted by quebec567

Hi! I'm a canadian student in america.

Please help me on division of radicals =)

1.) 6√18 ÷ 12√40

2.) 8√19 ÷ 4√38

1.) 6*sqrt(9)*sqrt(2)/[2*sqrt(10)]

18*sqrt(2)/[2*sqrt(10)]

(9*sqrt(2))/(sqrt(10))

(9*sqrt(5))/5

2.) (8*sqrt(19))/((4*sqrt(38))

(2*sqrt(2))/2

sqrt(2)
January 23rd 2007, 02:35 AM
AfterShock

Quote:

Originally Posted by quebec567

3.) 20√3 ÷ 5√ 3

4.) 42√ 6 ÷ 3√6

(20*sqrt(3))/(5*sqrt(3))

(4*sqrt(3))/(sqrt(3))

= 4

4.) (42*sqrt(6))/(3*sqrt(6))

(14*sqrt(6))/(sqrt(6))

= 14

And with that said, I have to get ready for work and such- I'll answer some of the other unanswered ones later.
January 23rd 2007, 03:39 AM
earboth

Quote:

Originally Posted by quebec567

Hi! I'm a canadian student in america.

Please help me on division of radicals =)

1.) 6√18 ÷ 12√40
2.) 8√19 ÷ 4√38
3.) 20√3 ÷ 5√ 3
4.) 42√ 6 ÷ 3√6
5.) -4√20 ÷ √2
6.) 1 / 2+√5
7.) 1 / 3-√11
8.) 3 / √3-1
9.) 5 / 2-√3
10.) 2 / 4-√14
thanks!

Hello,

$\frac{6\sqrt{18}}{12\sqrt{40}}=\frac{6 \cdot 3 \cdot \sqrt{2}}{12 \cdot 2 \cdot \sqrt{5}}=\frac{3\sqrt{2}}{4\sqrt{5}}=\frac{3\sqrt {10}}{20}$

$\frac{-4\sqrt{20}}{\sqrt{2}}=\frac{-4 \cdot 2 \sqrt{5}}{\sqrt{2}}=\frac{-8 \cdot \sqrt{10}}{2}=-4\sqrt{10}$

From #6 on I presume, that the denominators are sums(?):

$\frac{1}{2+\sqrt{5}} \cdot \frac{2-\sqrt{5}}{2-\sqrt{5}}=\frac{2-\sqrt{5}}{-1}=-2+\sqrt{5}$

$\frac{1}{3-\sqrt{11}} \cdot \frac{3+\sqrt{11}}{3+\sqrt{11}}=\frac{3+\sqrt{11}} {-2}$

$\frac{3}{\sqrt{3} - 1} \cdot \frac{\sqrt{3} + 1}{\sqrt{3} + 1}=\frac{3 \sqrt{3} + 3}{2}$

$\frac{5}{2-\sqrt{3}} \cdot \frac{2+\sqrt{3}}{2+\sqrt{3}}=\frac{10+5\sqrt{3}}{ 1}=10+5\sqrt{3}$

$\frac{2}{4-\sqrt{14}} \cdot \frac{4+\sqrt{14}}{4+\sqrt{14}}=\frac{8+2\sqrt{14} }{2}=4+\sqrt{14}$

EB
February 5th 2007, 08:13 PM
AlvinCY

Quote:

Originally Posted by earboth

Hello,

$\frac{6\sqrt{18}}{12\sqrt{40}}=\frac{6 \cdot 3 \cdot \sqrt{2}}{12 \cdot 2 \cdot \sqrt{5}}=\frac{3\sqrt{2}}{4\sqrt{5}}=\frac{3\sqrt {10}}{20}$

Don't you mean:

$\frac{6\sqrt{18}}{12\sqrt{40}} = \frac{6 \cdot 3 \cdot \sqrt{2}}{12 \cdot 2 \cdot \sqrt{10}} = \frac{3\sqrt{2}}{4\sqrt{10}} = \frac{3\sqrt{20}}{40} = \frac{3\sqrt{5}}{20}$
February 6th 2007, 12:28 PM
earboth

Quote:

Originally Posted by AlvinCY

Don't you mean:

$\frac{6\sqrt{18}}{12\sqrt{40}} = \frac{6 \cdot 3 \cdot \sqrt{2}}{12 \cdot 2 \cdot \sqrt{10}} = \frac{3\sqrt{2}}{4\sqrt{10}} = \frac{3\sqrt{20}}{40} = \frac{3\sqrt{5}}{20}$

Hi,

of course I meant that - thanks for correcting.

EB