If I have a "Dividing Radical Expressions" question and the denominator has something like 4+4 square roots of 5, would I need to times both of those numbers by the square root of 5?

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- anonymous

- chestercat

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- Owlfred

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- anonymous

Can you be more specific about what you're looking at? I can't quite understand your description.

- anonymous

multiply top and bottom by\[4-\sqrt{5}\]

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- anonymous

\[(4+\sqrt{5})(4-\sqrt{5})=16-5=11\]

- anonymous

so my equation is 3/ 4+4 square roots of 5. In order to solve the problem, there can be no radicals on the bottom half. Would I need to times both the 4 and the 4 square roots of 5 by the square root of 5?

- anonymous

Oh, as Satellite said. Multiply by the conjugate.
\[(a + \sqrt{b})(a - \sqrt{b}) = a^2 -b\]
So if you have \(a + \sqrt{b}\) you would multiply by \(a - \sqrt{b}\) and vice versa.

- anonymous

If you did that, you would get: \[3/(4+4\sqrt{5})=3 \sqrt{5}/(20+4\sqrt{5})\]
This does not cancel the square roots in the denominator so, instead, multiply the top and bottom by the conjugate of the denominator: \[4-4\sqrt{5}\]
This gives:
\[3/(4+4\sqrt{5})=3(4-4\sqrt{5})/(4+4\sqrt{5})(4-4\sqrt{5})=(12-12\sqrt{5})/(-64)\]
There are no square roots in the denominator of this expression so your problem is solved.

- anonymous

So what if the equation was that the alone 4 was negative

- anonymous

Wouldn't matter.

- anonymous

The part you change the sign on is the part with the radical.

- anonymous

sorry i did not read carefully. if it was \[4+4\sqrt{5}\] you multiply by \[4-4\sqrt{5}\]

- anonymous

When I use the conjugated form, do I need to change the negative 3 to a positive 3?

- anonymous

multiply by the 'conjugate'

- anonymous

conjugate of \[a+b\] is \[a-b\]

- anonymous

what if it was negitive a + b? then what would it be?

- anonymous

With square roots, the conjugate of \[a +\sqrt{b}\] is \[a-\sqrt{b}\]Basically, change the sign on the square root to find the conjugate.

- anonymous

the point being that \[(a+b)(a-b)=a^2-b^2\]
so for example \[(3+\sqrt{2})(3-\sqrt{2})=3^2)-(\sqrt{2})^2=9-2=7\]

- anonymous

or \[(\sqrt{5}-3)(\sqrt{5}+3)=(\sqrt{5})^2-3^2=5-9=-4\] etc

- anonymous

Yes, multiplying by a conjugate always cancels the square roots.

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