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anonymous

  • 5 years ago

need help finding 2 solutions for this equation.

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  1. anonymous
    • 5 years ago
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    \[y ^{2\over3}+5y ^{1\over3}-104=0\]

  2. dumbcow
    • 5 years ago
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    treat it just like a quadratic equation since (y^1/3)^2 = y^2/3 factor , what factors of -104 add up to 5

  3. anonymous
    • 5 years ago
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    i couldn't think of anything that does.

  4. dumbcow
    • 5 years ago
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    start with the obvious 52*2 then take half of 52 and double 2 26*4 13*8

  5. anonymous
    • 5 years ago
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    -13 and 8 work I guess

  6. dumbcow
    • 5 years ago
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    check the signs, they need to add up to positive 5

  7. anonymous
    • 5 years ago
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    13 and -8

  8. dumbcow
    • 5 years ago
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    ok so now (y^1/3 +13)(y^1/3 - 8) = 0 solve each part for y

  9. anonymous
    • 5 years ago
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    so 13- 1/3 and 8-1/3? Sorry, I know this is way more simple than I'm making it.

  10. anonymous
    • 5 years ago
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    i just don't know if it will end up not being a whole number this way? and what that would be exactly

  11. dumbcow
    • 5 years ago
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    thats ok y^1/3 = -13 to get y by itself i have to take each side to the 3rd power this way (y^1/3)^3 = y y = -13^3 also y = 8^3

  12. anonymous
    • 5 years ago
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    oh ok. so the answers can be to the third power like that?

  13. anonymous
    • 5 years ago
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    my online assignment wont let me type the answer to the third power. Do I solve more from there?

  14. dumbcow
    • 5 years ago
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    put it in your calculator to get a whole number

  15. anonymous
    • 5 years ago
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    one of them is 28561 and the other is4096..? kinda big numbers.

  16. dumbcow
    • 5 years ago
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    8^3 = 8*8*8 = 512 -13^3 = -13*-13*-13 = -2197

  17. anonymous
    • 5 years ago
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    ohho k, i didn't know the answers could be that large.

  18. dumbcow
    • 5 years ago
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    only because of the fractions in the exponents

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