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anonymous

  • one year ago

Find (x^2+1)/x if ...

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  1. anonymous
    • one year ago
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    \[\sqrt[3]{x^{2}}-3\sqrt[3]{x}+1=0\]

  2. mathmath333
    • one year ago
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    try to put \(x^{1/3}=a\) and solve quadratic

  3. anonymous
    • one year ago
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    i tried that but i could not get any result ??

  4. mathmath333
    • one year ago
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    that method looks cumbersome anyway

  5. triciaal
    • one year ago
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    |dw:1443258523188:dw|

  6. triciaal
    • one year ago
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    |dw:1443258789930:dw|

  7. anonymous
    • one year ago
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    but if a includes a square root, i can't find (x^2+1)/x as a reel number the answer to this question is 18

  8. triciaal
    • one year ago
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    |dw:1443259121075:dw|

  9. triciaal
    • one year ago
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    |dw:1443259344305:dw|

  10. imqwerty
    • one year ago
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    ok i got a solution we need to find-\[\frac{ x^2+1 }{ x } =x+\frac{ 1 }{ x}\] we have \[\sqrt[3]{x^2}-3\sqrt[3]{x}+1=0\] \[x^\left( \frac{ 1 }{ 3 } \right)=\alpha\]\[\alpha^2-3\alpha+1=0\] lets say that the roots are α1 and α2 so the product of roots is =c/a =1 \[\alpha_{1} \times \alpha_{2}=1\] \[\alpha_{1}=\frac{ 1 }{ \alpha_{2} }\]so the roots are reciprocal of each other so the two roots are α and 1/α and α+1/α=-b/a =3 \[\left( x+\frac{ 1 }{ x } \right)^2 =x^2 +\frac{ 1 }{ x^2} + 2\times x \times \frac{ 1 }{ x }\] we know α=x^2 and x^2 +1/x^2 =3 substitute u get- \[\left( x+\frac{ 1 }{ x } \right)^2=3+2 =>\left( x+\frac{ 1 }{ x} \right)=\sqrt{5}\]

  11. imqwerty
    • one year ago
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    if u get confused between those a, b and c then let me tell- we had a quadratic equation is α it was\[α^2-3α+1=0\]and the standard form of a quadratic equation is \[ax^2+bx+c=0\] and sum of roots=-b/a products of roots=c/a comparing the two equations u can see that a=1 b=-3 and c=1

  12. anonymous
    • one year ago
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    the answer is 18

  13. triciaal
    • one year ago
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    @strenesmee do you want to finish this to see if is indeed 18?

  14. imqwerty
    • one year ago
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    ^

  15. triciaal
    • one year ago
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    or wait for @imqwerty to repeat then continue

  16. anonymous
    • one year ago
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    s/he got square root 5

  17. imqwerty
    • one year ago
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    umm can u help me find any mistake which i made >.< #he

  18. imqwerty
    • one year ago
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    u sure its 18?

  19. imqwerty
    • one year ago
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    well if x+1/x=18 then \[\left( x+\frac{ 1 }{ x } \right)^2=x^2+\frac{ 1 }{ x^2 }+2\times x \times \frac{ 1 }{ x }\]\[18^2 -2=x^2+\frac{ 1 }{ x^2 } =322\]haahah and x^2 and 1/x^2 are the roots of α^2-3α+1=0 and the sum of roots which we got is 322 but from the equation its clear that the sum is 3 and not 322 :P so 18 cannot be the answer answer is root(5)

  20. triciaal
    • one year ago
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    strong possibility not eighteen but rather x = 1; x = 8

  21. triciaal
    • one year ago
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    |dw:1443261857218:dw|

  22. triciaal
    • one year ago
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    |dw:1443261948201:dw|

  23. triciaal
    • one year ago
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    |dw:1443262059639:dw|

  24. triciaal
    • one year ago
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    may have a sign error or use positive only

  25. Jhannybean
    • one year ago
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    \[\begin{align} \\ \qquad a^2-3a+1 &= (a^2-3a)+1 \\ &= \left(a^2-3a+\frac{9}{4}\right)+1-\frac{9}{4} \\ &= \left(a-\frac{3}{2}\right)^2 -\frac{5}{4} \end{align}\] then \(a=x^{1/3}\) right?...

  26. imqwerty
    • one year ago
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    yes

  27. Jhannybean
    • one year ago
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    \[\left(a-\frac{3}{2}\right)^2 -\frac{5}{4} =0\]\[a-\frac{3}{2} =\pm \frac{\sqrt{5}}{2}\]\[a=\frac{3}{2}\pm \frac{\sqrt{5}}{2}\] Would the substitution work AFTER solving for a?.... :D Haha

  28. Jhannybean
    • one year ago
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    Yeah.... looking at @triciaal 's work maybe not lol.

  29. imqwerty
    • one year ago
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    cx

  30. anonymous
    • one year ago
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  31. anonymous
    • one year ago
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    i have finally learned the correct answer @imqwerty @triciaal

  32. imqwerty
    • one year ago
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    >.< i got my mistake i took the roots x^2 and 1/x^2 but that should have been x^3 and 1/x^3 >.< ;-; nvm thanks

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