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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sqrt[3]{x^{2}}3\sqrt[3]{x}+1=0\]

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1try to put \(x^{1/3}=a\) and solve quadratic

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i tried that but i could not get any result ??

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1that method looks cumbersome anyway

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443258523188:dw

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443258789930:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but 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

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443259121075:dw

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443259344305:dw

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0ok i got a solution we need to find\[\frac{ x^2+1 }{ x } =x+\frac{ 1 }{ x}\] we have \[\sqrt[3]{x^2}3\sqrt[3]{x}+1=0\] \[x^\left( \frac{ 1 }{ 3 } \right)=\alpha\]\[\alpha^23\alpha+1=0\] 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 \[\left( x+\frac{ 1 }{ x } \right)^2=3+2 =>\left( x+\frac{ 1 }{ x} \right)=\sqrt{5}\]

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0if u get confused between those a, b and c then let me tell we had a quadratic equation is α it was\[α^23α+1=0\]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

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0@strenesmee do you want to finish this to see if is indeed 18?

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0or wait for @imqwerty to repeat then continue

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0s/he got square root 5

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0umm can u help me find any mistake which i made >.< #he

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0well 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^23α+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)

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0strong possibility not eighteen but rather x = 1; x = 8

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443261857218:dw

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443261948201:dw

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443262059639:dw

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0may have a sign error or use positive only

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\begin{align} \\ \qquad a^23a+1 &= (a^23a)+1 \\ &= \left(a^23a+\frac{9}{4}\right)+1\frac{9}{4} \\ &= \left(a\frac{3}{2}\right)^2 \frac{5}{4} \end{align}\] then \(a=x^{1/3}\) right?...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\left(a\frac{3}{2}\right)^2 \frac{5}{4} =0\]\[a\frac{3}{2} =\pm \frac{\sqrt{5}}{2}\]\[a=\frac{3}{2}\pm \frac{\sqrt{5}}{2}\] Would the substitution work AFTER solving for a?.... :D Haha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah.... looking at @triciaal 's work maybe not lol.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i have finally learned the correct answer @imqwerty @triciaal

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0>.< 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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