anonymous
  • anonymous
I need help on simplifying radicals, so could someone explain to me how to solve this? 2√16a^4b^2
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.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Astrophysics
  • Astrophysics
\[2\sqrt{16a^4b^2}\]?
anonymous
  • anonymous
Yes.
Astrophysics
  • Astrophysics
So since it's being multiplied you can split it all up as such \[2\sqrt{16}\sqrt{a^4}\sqrt{b^2}\] can you simplify from there?

Looking for something else?

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

More answers

anonymous
  • anonymous
How would you simplify from 16? Because don't you need one number that is a perfect square and the other one not? The only thing I can think of is 4x4 but they're both perfect squares...
Astrophysics
  • Astrophysics
\[\sqrt{16}=4\]
Astrophysics
  • Astrophysics
\[\sqrt{x} = x^{1/2}\] \[\sqrt{a^4}=(a^4)^{1/2}=a^{4/2}=a^{2}\] do the same with \[\sqrt{b^2}\]
Astrophysics
  • Astrophysics
Also remember \[4^2 = 16\]
anonymous
  • anonymous
Okay, so would it be \[8\sqrt{a ^{2}}\]
anonymous
  • anonymous
Because the "b" cancels out because it's b squared?
Astrophysics
  • Astrophysics
No, look at what I wrote above, \[\sqrt{x} = x^{1/2}\] this is the same thing, just exponent. \[\sqrt{b^2} = (b^2)^{1/2} \] and for your other question \[\sqrt{16} = \sqrt{4 \times 4} = \sqrt{4}\sqrt{4} = 2 \times 2 = 4\]
anonymous
  • anonymous
I am a moron when it comes to math, so bear with me. Soooo sorry. :/
Astrophysics
  • Astrophysics
It's ok :P
anonymous
  • anonymous
So it would just be b? Gosh, exponents really confuse me.
Astrophysics
  • Astrophysics
Yup perfect! Here is a little guide that a user iambatman made for exponents, it may help clear some things up |dw:1439179339225:dw|
anonymous
  • anonymous
\[8b \sqrt{a ^{2}}\]
anonymous
  • anonymous
would it be this?
Astrophysics
  • Astrophysics
Not quite, \[8a^2b\]
Astrophysics
  • Astrophysics
Remember it was \[\sqrt{a^4}\]
anonymous
  • anonymous
So the a goes on the other side of the radical when simplified?
Astrophysics
  • Astrophysics
I don't quite follow
Astrophysics
  • Astrophysics
Oh, there is no point to put square root sign, a^2 is simplied
anonymous
  • anonymous
It goes outside of the radical/square root when it got simplified from \[a^{4} \to a ^{2}\]?
Astrophysics
  • Astrophysics
\[\huge \sqrt{a^4}=(a^4)^{1/2}=a^{4/2}=a^{2}\]
Astrophysics
  • Astrophysics
Does that make sense?
anonymous
  • anonymous
Yes, it does. I understand now. Thank you so much! I had a horrible teacher when I took Algebra.. I practically had to teach myself everything, so I struggle a lot. Thanks again!
Astrophysics
  • Astrophysics
When we struggle we learn! And your welcome :)

Looking for something else?

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