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anonymous

  • one year ago

I need help on simplifying radicals, so could someone explain to me how to solve this? 2√16a^4b^2

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  1. Astrophysics
    • one year ago
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    \[2\sqrt{16a^4b^2}\]?

  2. anonymous
    • one year ago
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    Yes.

  3. Astrophysics
    • one year ago
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    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?

  4. anonymous
    • one year ago
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    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...

  5. Astrophysics
    • one year ago
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    \[\sqrt{16}=4\]

  6. Astrophysics
    • one year ago
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    \[\sqrt{x} = x^{1/2}\] \[\sqrt{a^4}=(a^4)^{1/2}=a^{4/2}=a^{2}\] do the same with \[\sqrt{b^2}\]

  7. Astrophysics
    • one year ago
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    Also remember \[4^2 = 16\]

  8. anonymous
    • one year ago
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    Okay, so would it be \[8\sqrt{a ^{2}}\]

  9. anonymous
    • one year ago
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    Because the "b" cancels out because it's b squared?

  10. Astrophysics
    • one year ago
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    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 \[\sqrt{16} = \sqrt{4 \times 4} = \sqrt{4}\sqrt{4} = 2 \times 2 = 4\]

  11. anonymous
    • one year ago
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    I am a moron when it comes to math, so bear with me. Soooo sorry. :/

  12. Astrophysics
    • one year ago
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    It's ok :P

  13. anonymous
    • one year ago
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    So it would just be b? Gosh, exponents really confuse me.

  14. Astrophysics
    • one year ago
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    Yup perfect! Here is a little guide that a user iambatman made for exponents, it may help clear some things up |dw:1439179339225:dw|

  15. anonymous
    • one year ago
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    \[8b \sqrt{a ^{2}}\]

  16. anonymous
    • one year ago
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    would it be this?

  17. Astrophysics
    • one year ago
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    Not quite, \[8a^2b\]

  18. Astrophysics
    • one year ago
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    Remember it was \[\sqrt{a^4}\]

  19. anonymous
    • one year ago
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    So the a goes on the other side of the radical when simplified?

  20. Astrophysics
    • one year ago
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    I don't quite follow

  21. Astrophysics
    • one year ago
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    Oh, there is no point to put square root sign, a^2 is simplied

  22. anonymous
    • one year ago
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    It goes outside of the radical/square root when it got simplified from \[a^{4} \to a ^{2}\]?

  23. Astrophysics
    • one year ago
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    \[\huge \sqrt{a^4}=(a^4)^{1/2}=a^{4/2}=a^{2}\]

  24. Astrophysics
    • one year ago
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    Does that make sense?

  25. anonymous
    • one year ago
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    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!

  26. Astrophysics
    • one year ago
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    When we struggle we learn! And your welcome :)

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