anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Owlfred
  • Owlfred
Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!
anonymous
  • anonymous
Can you be more specific about what you're looking at? I can't quite understand your description.
anonymous
  • anonymous
multiply top and bottom by\[4-\sqrt{5}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
\[(4+\sqrt{5})(4-\sqrt{5})=16-5=11\]
anonymous
  • 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
  • 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
  • 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
  • anonymous
So what if the equation was that the alone 4 was negative
anonymous
  • anonymous
Wouldn't matter.
anonymous
  • anonymous
The part you change the sign on is the part with the radical.
anonymous
  • anonymous
sorry i did not read carefully. if it was \[4+4\sqrt{5}\] you multiply by \[4-4\sqrt{5}\]
anonymous
  • anonymous
When I use the conjugated form, do I need to change the negative 3 to a positive 3?
anonymous
  • anonymous
multiply by the 'conjugate'
anonymous
  • anonymous
conjugate of \[a+b\] is \[a-b\]
anonymous
  • anonymous
what if it was negitive a + b? then what would it be?
anonymous
  • 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
  • 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
  • anonymous
or \[(\sqrt{5}-3)(\sqrt{5}+3)=(\sqrt{5})^2-3^2=5-9=-4\] etc
anonymous
  • anonymous
Yes, multiplying by a conjugate always cancels the square roots.

Looking for something else?

Not the answer you are looking for? Search for more explanations.