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

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

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[\sqrt[3]{x^{2}}-3\sqrt[3]{x}+1=0\]
try to put \(x^{1/3}=a\) and solve quadratic
i tried that but i could not get any result ??

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Other answers:

that method looks cumbersome anyway
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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
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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- \[\left( x+\frac{ 1 }{ x } \right)^2=3+2 =>\left( x+\frac{ 1 }{ x} \right)=\sqrt{5}\]
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
the answer is 18
@strenesmee do you want to finish this to see if is indeed 18?
^
or wait for @imqwerty to repeat then continue
s/he got square root 5
umm can u help me find any mistake which i made >.< #he
u sure its 18?
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)
strong possibility not eighteen but rather x = 1; x = 8
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may have a sign error or use positive only
\[\begin{align} \\ \qquad a^2-3a+1 &= (a^2-3a)+1 \\ &= \left(a^2-3a+\frac{9}{4}\right)+1-\frac{9}{4} \\ &= \left(a-\frac{3}{2}\right)^2 -\frac{5}{4} \end{align}\] then \(a=x^{1/3}\) right?...
yes
\[\left(a-\frac{3}{2}\right)^2 -\frac{5}{4} =0\]\[a-\frac{3}{2} =\pm \frac{\sqrt{5}}{2}\]\[a=\frac{3}{2}\pm \frac{\sqrt{5}}{2}\] Would the substitution work AFTER solving for a?.... :D Haha
Yeah.... looking at @triciaal 's work maybe not lol.
cx
i have finally learned the correct answer @imqwerty @triciaal
>.< 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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