anonymous
  • anonymous
Find (x^2+1)/x if ...
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\sqrt[3]{x^{2}}-3\sqrt[3]{x}+1=0\]
mathmath333
  • mathmath333
try to put \(x^{1/3}=a\) and solve quadratic
anonymous
  • anonymous
i tried that but i could not get any result ??

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

mathmath333
  • mathmath333
that method looks cumbersome anyway
triciaal
  • triciaal
|dw:1443258523188:dw|
triciaal
  • triciaal
|dw:1443258789930:dw|
anonymous
  • 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
triciaal
  • triciaal
|dw:1443259121075:dw|
triciaal
  • triciaal
|dw:1443259344305:dw|
imqwerty
  • 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- \[\left( x+\frac{ 1 }{ x } \right)^2=3+2 =>\left( x+\frac{ 1 }{ x} \right)=\sqrt{5}\]
imqwerty
  • 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
anonymous
  • anonymous
the answer is 18
triciaal
  • triciaal
@strenesmee do you want to finish this to see if is indeed 18?
imqwerty
  • imqwerty
^
triciaal
  • triciaal
or wait for @imqwerty to repeat then continue
anonymous
  • anonymous
s/he got square root 5
imqwerty
  • imqwerty
umm can u help me find any mistake which i made >.< #he
imqwerty
  • imqwerty
u sure its 18?
imqwerty
  • 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)
triciaal
  • triciaal
strong possibility not eighteen but rather x = 1; x = 8
triciaal
  • triciaal
|dw:1443261857218:dw|
triciaal
  • triciaal
|dw:1443261948201:dw|
triciaal
  • triciaal
|dw:1443262059639:dw|
triciaal
  • triciaal
may have a sign error or use positive only
Jhannybean
  • Jhannybean
\[\begin{align} \\ \qquad a^2-3a+1 &= (a^2-3a)+1 \\ &= \left(a^2-3a+\frac{9}{4}\right)+1-\frac{9}{4} \\ &= \left(a-\frac{3}{2}\right)^2 -\frac{5}{4} \end{align}\] then \(a=x^{1/3}\) right?...
imqwerty
  • imqwerty
yes
Jhannybean
  • Jhannybean
\[\left(a-\frac{3}{2}\right)^2 -\frac{5}{4} =0\]\[a-\frac{3}{2} =\pm \frac{\sqrt{5}}{2}\]\[a=\frac{3}{2}\pm \frac{\sqrt{5}}{2}\] Would the substitution work AFTER solving for a?.... :D Haha
Jhannybean
  • Jhannybean
Yeah.... looking at @triciaal 's work maybe not lol.
imqwerty
  • imqwerty
cx
anonymous
  • anonymous
anonymous
  • anonymous
i have finally learned the correct answer @imqwerty @triciaal
imqwerty
  • 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.