## anonymous 5 years ago 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?

1. Owlfred

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2. anonymous

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

3. anonymous

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

4. anonymous

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

5. 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?

6. 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.

7. 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.

8. anonymous

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

9. anonymous

Wouldn't matter.

10. anonymous

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

11. anonymous

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

12. anonymous

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

13. anonymous

multiply by the 'conjugate'

14. anonymous

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

15. anonymous

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

16. 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.

17. anonymous

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

18. anonymous

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

19. anonymous

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