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anonymous

  • one year ago

Medal + fan Simplify the expression. x times the square root of the quantity 8 x cubed, plus x times the square root of 2, plus four times the square root of the quantity 2x

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  1. anonymous
    • one year ago
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    \[x \sqrt{8x ^{3}} + x \sqrt{2} + 4\sqrt{2x}\]

  2. anonymous
    • one year ago
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    I'm usually good at math but this question is a little confusing.... So far I think the answer is 2 because yo can simplify it to get 2x^2sqrt(2x)... and so on

  3. anonymous
    • one year ago
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    \[2\sqrt{2} \times x \times x ^{3/2} + x \times \sqrt{2} + 4 \sqrt{2} \times x^{1/2} \]

  4. anonymous
    • one year ago
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    To write it into its raw exponential form.

  5. anonymous
    • one year ago
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    why did you put fractions in the radical?

  6. anonymous
    • one year ago
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    By putting the exponents in their fraction forms, you can clearly see their combined values.

  7. anonymous
    • one year ago
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    This will allow you to factor out an exponential of x, through subtraction of the exponent.

  8. anonymous
    • one year ago
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    As you can see in this question, the lowest "factorable" term is \[\sqrt{x}\]

  9. anonymous
    • one year ago
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    I'm only in ninth grade, I don't think this is necessary.. besides, I haven't learned this yet.

  10. anonymous
    • one year ago
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    I'm not looking for the lowest factorable term.... I'm looking for the leading coefficient when it's in simplified form

  11. anonymous
    • one year ago
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    But it's factored form looks something like this: \[\sqrt{2} \sqrt{x} (4+\sqrt{x}+2 x^2)\]

  12. anonymous
    • one year ago
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    To do that, you need write out the terms in full and subtract the exponents when factoring.

  13. anonymous
    • one year ago
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    I think we're looking for the simplified form not the factored form... \[2x^2 \sqrt{2x} + x \sqrt{2} + 4 \sqrt{2x}\]

  14. anonymous
    • one year ago
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    Is this fully simplified?

  15. anonymous
    • one year ago
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    Well, I suppose you can't go any further to "simplify" it without factoring.

  16. anonymous
    • one year ago
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    ok, so we agree that the leading coefficient is 2?

  17. anonymous
    • one year ago
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    Okay, thanks, I got it right!

  18. anonymous
    • one year ago
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    The leading coefficient of the entire term is not 2, its \[2\sqrt{2}\]

  19. anonymous
    • one year ago
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    100% on my quiz! woot!

  20. anonymous
    • one year ago
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    Great to hear!

  21. anonymous
    • one year ago
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    Well, I guess it was the answer they were looking for!

  22. anonymous
    • one year ago
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    As promised... medal and fan....

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