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anonymous

  • 5 years ago

3x^1/2-8x^1/4-3=0 Can you show me how to solve?

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  1. anonymous
    • 5 years ago
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    take x^(1/4)=y

  2. anonymous
    • 5 years ago
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    solve the quadratic in y.

  3. anonymous
    • 5 years ago
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    You are to set up as a substitution problem and solve.

  4. anonymous
    • 5 years ago
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    yes. solve 3y^2-8y-3=0. lastly substitute x^(1/4) for y.

  5. anonymous
    • 5 years ago
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    I came up with a substituion of u2-8u-3=0 and came up with 4+-square root 19 but I am not understanding how to substitute the original u=x1/4 and come up with the right answer

  6. anonymous
    • 5 years ago
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    its not u^2-8u-3=0 but 3u^2-8u-3=0. if u came with something like u=4+sqrt(19). then x=u^4=(4+sqrt(19))^4. Thats it. :)

  7. anonymous
    • 5 years ago
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    Not exactly. You have to substitute back in the x 1/4 for the u. That is where I am stuck and not coming up with the correct answer.

  8. anonymous
    • 5 years ago
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    explain where you are stuck.

  9. anonymous
    • 5 years ago
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    I solved the quadratics and got u=3,-1/3. then x=3^4=81 or x=(-1/3)^4=1/81.

  10. anonymous
    • 5 years ago
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    But x=1/81 does not satisfy the orginal eqn . so our ans is 81

  11. anonymous
    • 5 years ago
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    ok. on the original problem I made u=x1/4. At this point the equation should have been 3u^2-8u-3=0. I missed that, you were correct. At that point I plugged it in the quadratic equation. Bottom line I am coming up with 9, 1. So my substituion back should be x1/4=9 and x1/4=1. How do you solve these?

  12. anonymous
    • 5 years ago
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    why? x=9^4 or x=1 with ur values.

  13. anonymous
    • 5 years ago
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    it is x 1/4 = 9, x 1/4 =1

  14. anonymous
    • 5 years ago
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    what is x1/4? Isnt it x^(1/4)?

  15. anonymous
    • 5 years ago
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    Im sorry it is x^1/4 =9 and x^1/4 =1

  16. anonymous
    • 5 years ago
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    let u=x^1/4 so that u^2 =x^1/2 3u^2 - 8u -3=0 using the quadratic formula u= [-b+ sqrt(b^2-4ac)]/2a

  17. anonymous
    • 5 years ago
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    so where are u getting the problem? Raise both sides of the equality (x^1/4=9) to power 4. and u get x=9^4. Is this ur problem?

  18. anonymous
    • 5 years ago
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    Yep Mark. That's what I did.

  19. anonymous
    • 5 years ago
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    Suabhik, yes. that is my problem.

  20. anonymous
    • 5 years ago
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    is it x = 6561 and x=1?

  21. anonymous
    • 5 years ago
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    yups bastax.

  22. anonymous
    • 5 years ago
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    Well alrighty then. Just talking it through really helped. Thanks saubhik!

  23. anonymous
    • 5 years ago
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    hope u r clear with it. :)

  24. anonymous
    • 5 years ago
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    i better be. I have a test in an hour!

  25. anonymous
    • 5 years ago
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    best of luck!

  26. anonymous
    • 5 years ago
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    ok u= [-(-8)+ sqrt((-8)^2-4(3)(-3))]/2*3 u=[8+ sqrt(64+36)]/6 u={8+sqrt(100)/9 u= [8+10]/6 u= 18/6 =3 also the other u is -1/3

  27. anonymous
    • 5 years ago
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    Back again on this same question. Book says answer should be 81 and the extraneous is 1/81. How did they get there?

  28. anonymous
    • 5 years ago
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    u=x^1/4 =3 x=(3)^4=81

  29. anonymous
    • 5 years ago
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    u=(-1/3)^4= 1/81

  30. anonymous
    • 5 years ago
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    hope u get the procedure of getting it

  31. anonymous
    • 5 years ago
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    and then by using the quadratic formula

  32. anonymous
    • 5 years ago
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    I got down to x^1/4=9 and x^1/4 =1. Was this correct? Then how did you come up with the 81, 1/81?

  33. anonymous
    • 5 years ago
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    No this is not correct. Please check ur equation.

  34. anonymous
    • 5 years ago
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    read the one i wrote on the top one and the second part

  35. anonymous
    • 5 years ago
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    let u=x^1/4 so that u^2 =x^1/2 3u^2 - 8u -3=0 using the quadratic formula u= [-b+ sqrt(b^2-4ac)]/2a u= [-(-8)+ sqrt((-8)^2-4(3)(-3))]/2*3 u=[8+ sqrt(64+36)]/6 u={8+sqrt(100)/9 u= [8+10]/6 u= 18/6 =3 also the other u is -1/3

  36. anonymous
    • 5 years ago
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    ok. I can see how you got the 3, -1/3. I had made a mistake earlier in the problem.

  37. anonymous
    • 5 years ago
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    u=x^1/4 =3 x=(3)^4=81 u=(-1/3)^4= 1/81

  38. anonymous
    • 5 years ago
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    ok good luck

  39. anonymous
    • 5 years ago
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    Got it!!!! Thank you so much Mark!

  40. anonymous
    • 5 years ago
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    ok welcome,,,,, saubnik also got it in his answer...

  41. anonymous
    • 5 years ago
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    ok thnx....good luck with our exam

  42. anonymous
    • 5 years ago
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    your exam lol

  43. anonymous
    • 5 years ago
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    Yep. I saw you both had it! Thanks again!

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