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anonymous

  • 4 years ago

i need help solving this: x^(2/3)-2x^(1/3)=15. And the intrsutions say explain and demonstrate how to use substitution to solve the equation. well i don't understand this at all. please help!

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  1. anonymous
    • 4 years ago
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    treat x^(2/3) as x^(1/3)^2

  2. anonymous
    • 4 years ago
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    ok put \[z=x^{\frac{1}{3}}\]

  3. anonymous
    • 4 years ago
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    then factor as x^2-2x-15=0, and substitute x^1/3 back in for x

  4. anonymous
    • 4 years ago
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    then by the laws of exponents \[z^2=(x^{\frac{1}{3}})^2=x^{\frac{2}{3}}\]

  5. anonymous
    • 4 years ago
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    so by replacing \[x^{\frac{1}{3}}\] by \[z\] you get an equation that looks like \[z^2-2z=15\]

  6. anonymous
    • 4 years ago
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    so the answer should be 5?

  7. anonymous
    • 4 years ago
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    there are two answers

  8. anonymous
    • 4 years ago
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    and you can solve this for z by writing \[z^2-2z-15=0\] \[(z-5)(z+3)=0\] so \[z=5\] or \[z=-3\]

  9. anonymous
    • 4 years ago
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    that is what i was getting 5 or -3 but then when i try to substitute back in i do not get anything that looks right. i absolutely think that i am doing something wrong still

  10. anonymous
    • 4 years ago
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    but now we have to remember that \[z=x^{\frac{1}{3}}\]so that means that \[x^{\frac{1}{3}}=5\] or \[x^{\frac{1}{3}}=-3\]

  11. anonymous
    • 4 years ago
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    and so \[x=(5)^2=125\] or \[x=(-3)^3=-27\]

  12. anonymous
    • 4 years ago
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    so i substitute back in -25 or 127?

  13. anonymous
    • 4 years ago
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    i mean -27 or 125

  14. anonymous
    • 4 years ago
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    you substitute x^(1/3) back in for z

  15. anonymous
    • 4 years ago
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    x^(1/3)=5 x^(1/3)=-3

  16. anonymous
    • 4 years ago
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    then cube both sides of each equation to find x

  17. anonymous
    • 4 years ago
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    (x^(1/3))^3=5^3 x=125 (x^(1/3))^3=-3^3 x=-27

  18. anonymous
    • 4 years ago
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    so both are good solutions? im trying to follow sorry, i am just very bad at this

  19. anonymous
    • 4 years ago
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    so both are good solutions? im trying to follow sorry, i am just very bad at this

  20. anonymous
    • 4 years ago
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    Yes both are good solutions because they are in the domain of the original function.

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