And then I said...

Algebra Level 4

Find the value of ${ x }^{ 2 }+\cfrac { 1 }{ { x }^{ 2 } }$ . Given that $0<x$ and $x+\cfrac { 1 }{ x } =1$ .

\cfrac { -\sqrt { 3 } }{ 2 } +\cfrac { 1 }{ 2 }

none of these

\cfrac { \sqrt { -3 } }{ 2 } +\cfrac { 1 }{ 2 }

-1

2 solutions

Soumo Mukherjee
Jan 15, 2015

For any $x>0$ we have $\displaystyle \cfrac { \left( x+\cfrac { 1 }{ x } \right) }{ 2 } \ge \sqrt { \left( x\times \cfrac { 1 }{ x } \right) }$ i.e $AM\ge GM$ . Which means that $\left( x+\cfrac { 1 }{ x } \right) \ge 2$ .

But in the question we have $x>0$ AND ${ \left( x+\cfrac { 1 }{ x } \right) }=1$ , which is impossible. Hence none of the answers are correct.

(x+1/x)^2=1 implies x^2+(1/x)^2=-1 ;so there does exist a complex number that satisfy your conditions. You DID NOT rule the complex value in your question.

aaron paul - 6 years, 4 months ago

While I did understand it as real $x$ because of $x > 0$ (which is undefined for complex numbers), I agree that it should be clarified.

Jakub Šafin - 6 years, 4 months ago

Please notice that in the question I have mentioned $x>0$ . If you have more doubts, feel free to clarify them. :)

Soumo Mukherjee - 6 years, 4 months ago

how? please explain

Sanjoy Roy - 6 years, 4 months ago

for any $x>0$ we have $\displaystyle { \left( x+\cfrac { 1 }{ x } \right) }/{ 2 }\ge \sqrt { \left( x\times \cfrac { 1 }{ x } \right) } \\ \Rightarrow \left( x+\cfrac { 1 }{ x } \right) \ge 2$ .

Our question sates that $x>0$ and $\displaystyle { \left( x+\cfrac { 1 }{ x } \right) }=1$ . Which is impossible.

Soumo Mukherjee - 6 years, 4 months ago

William Isoroku
Jan 17, 2015

$x+\frac{1}{x}=1\longrightarrow x^2-x+1=0$ , which has imaginary roots. Therefore $x$ doesn't fit the condition.

And then I said...

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