Question

$2(x+ \frac{1}{x})^{2}-7(x + \frac{1}{x} ) + 5 = 0 \\ solve \: this$
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Answer 1

Step-by-step explanation:

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Suppose x+1x=1

For every non-zero real x , x and 1x are either both positive or they are both negative.

Since their sum is positive (i.e. 1), we can conclude that they would have to be positive.

Now, for any x>0 we have (x−1)2>0 .

x2−2x+1>0

x2+1>2x

As x is positive, we can divide by it without flipping the sign.

x+1x>2

The LHS of this inequality is the same as the LHS of our premise. So we can plug in the premise, getting:

1>2

Therefore, by reductio ad absurdum x is not a real number,

Let us then consider complex x=r⋅eiϕ=rcosϕ+irsinϕ

Then 1x=1r⋅e−iϕ=1rcosϕ−1risinϕ .

So, x+1x=cosϕ(r+1r)+isinϕ(r−1r)=1+0i

therefore,

(r+1r)cosϕ=1 and

(r−1r)sinϕ=0

From the latter we get either sinϕ=0 , which means x is a real number - this option has been disproven - or r−1r=0 . Given that a complex number’s modulus is a non-negative real, the only r satisfying the equation is 1.

Then let us consider r=1 . We get (1+1)cosϕ=1 so cosϕ=12 .

This means ϕ=π3 or ϕ=−π3.

These are not truly different ways. For x=eiπ3,1x=e−iπ3 The other option has them the other way round.

So, in this option x2+1x2=e2iπ3+e−2iπ3=2cos(2π3)=−1.

Therefore, if x+1x=1 then x2+1x2=−1 and x is a complex number.

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