## anonymous one year ago Find (x^2+1)/x if ...

1. anonymous

$\sqrt[3]{x^{2}}-3\sqrt[3]{x}+1=0$

2. mathmath333

try to put $$x^{1/3}=a$$ and solve quadratic

3. anonymous

i tried that but i could not get any result ??

4. mathmath333

that method looks cumbersome anyway

5. triciaal

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6. triciaal

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7. anonymous

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

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9. triciaal

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10. imqwerty

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

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

13. triciaal

@strenesmee do you want to finish this to see if is indeed 18?

14. imqwerty

^

15. triciaal

or wait for @imqwerty to repeat then continue

16. anonymous

s/he got square root 5

17. imqwerty

umm can u help me find any mistake which i made >.< #he

18. imqwerty

u sure its 18?

19. imqwerty

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

strong possibility not eighteen but rather x = 1; x = 8

21. triciaal

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22. triciaal

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23. triciaal

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24. triciaal

may have a sign error or use positive only

25. anonymous

\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

yes

27. anonymous

$\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. anonymous

Yeah.... looking at @triciaal 's work maybe not lol.

29. imqwerty

cx

30. anonymous

31. anonymous

i have finally learned the correct answer @imqwerty @triciaal

32. imqwerty

>.< 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