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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owlfred · Answer

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anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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 · Answer

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 · Answer

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 · Answer

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

anonymous · Answer

Wouldn't matter.

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

multiply by the 'conjugate'

anonymous · Answer

conjugate of $$a+b$$ is $$a-b$$

anonymous · Answer

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

anonymous · Answer

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 · Answer

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 · Answer

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

anonymous · Answer

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