A More Difficult Number-Plus-Reciprocal Problem

I didn't realize what's so difficult here! :/ Am I missing something?

$x+\dfrac{1}{x}=2 ~~~\Longrightarrow ~ x^2-2x+1=0 ~~~\Longrightarrow ~ (x-1)^2=0 ~~~\Longrightarrow ~ x=1$

$\therefore~~~ 1+1+1-1+1-1+2=\fbox{4}$

And how do we apply the fact that $2048=2^{11}$ ?

Since we don't exactly see how in the world we would get to these exponents, we start small. Squaring both sides of $x + \frac 1x = 2$ gives $x^2 + 2 + \frac{1}{x^2} = 4 \implies x^2 + \frac 1 {x^2} = 2 .$ Then we see that for any $n$, $x^n+ \frac 1{x^n} = 2 \implies x^{2n} + \frac 1{x^{2n}}$ . So we see that $x^{2048} + \frac{1}{x^{2048}} = 2$ . Now we have these other terms to deal with. Well, we can use $\left( x + \frac 1x \right) \left(x^{2048} + \frac{1}{x^{2048}} \right)$ However, this would not give us the negative terms that we need. So we need to find $x - \frac 1x$ . Since $x^2 + \frac{1}{x^2}= 2$ , we have $x^2 -2 + \frac{1}{x^2} = 0 \implies \left(x - \frac{1}{x}\right) ^2 = 0 \implies x - \frac 1x = 0.$ So this eliminates $\left( x - \frac 1x \right) \left(x^{2048} + \frac{1}{x^{2048}} \right)$ and we are left with $2 + 2 =4$ .

Note that $x=1$ fits the original equation; therefore we can see that the expression we want to find the value of is simply $1+1+1-1+1-1+2=\boxed{4}$ and we are done.

There is a solution by algebraic manipulation, but a more straightforward way is to solve the given hypothesis. As $x=0$ does not solve the equation, we may assume that $x\neq 0$ , and therefore multiply the equation by $x$ , solving the quadratic inequality gives $x=1$ . Substitute this into the expression and get the answer $\textbf{4}$ . Alternatively use the inequality $x+\frac{1}{x}\geq 2$ which is true for nonzero real $x$ , where equality occurs if and only if $x=1$ to get $x=1$ . In this case there is a gap that $x$ might be imaginary (not real). Assume so. Since $\frac{1}{x}=\frac{\overline{x}}{|x|}$ , for $x+\frac{1}{x}$ to have a zero imaginary part, $|x|=1$ , but then $x\neq 1$ by our assumption, $x$ 's real part is strictly less than $1$ , which makes $x+\frac{1}{x}$ to have a real part of strictly less than $2$ , contradiction.

Since we have : x + 1/x =2 , and in the equation we have two odd exponential value and one even, we can find the value of x^2 + 1/(x^2) for even and also x^3 + 1/(x^3) for the odd one, by this we can find the total value. Even value is 0 and odd value is 2

X+1/x=2 factor x(1+1/1)=2 simplify x(2)=2 isolate (x) x=2/2 solution x=1 plug in x=1 to original equation to get, 1+1+1-1+1-1+2=4

X^2-2x-1=0 Using quadratic. X=1 Substitute. 1+1/1+1-1/1+1/1-1+2=4

If you rearrange the basic information given we know: x + 1/x = 2 So: x^2 + 1 = 2x x^2 - 2x + 1 = 0 ( x - 1 )( x - 1) = 0 Therefore: x = 1 Now we can apply this to the question: The first section gives 1 + 1/1 which is 2 The next section gives: 1-1 which is 0 The third section gives: 1-1 which is also 0 ( similar to second section) The last section is 2 If we add all the sections we get ; 2 + 0 + 0 + 2 which gives 4

1 to the power of any whole number is 1. All you are doing is adding and subtracting 1.

Let x=1, hence question solved

So firstly I differentiated x+1/x = 2 and got x^2=1 , so x =+-1 , but - 1dont satisfy the equation x+1/x=2 so I get x=+1 , now put x=1 in long equation to get that reduced to the value 4 .

x+(1/x)=2 => x(1+(1/1)=2 => 2x=2 => x=1 1+1+1-1+1-1+2 = 4

Looking at reciprocals each denom and numer can be canceled bc of its properties in the problem. Leaving it equaling 2 and adding 2 at the end to equal 4

You literally multiply the equation by x solve for x and then plug in the value into the expression

Wow, so many complicated ways when all I did was switching + to - and - to +... Then just apply -x-1/x = -2 and you can even calculate it in your head : )

That wasnt difficult . Everyone would have solved that in first attempt !!

x+1/x=2 has only one solution . x=1. 1^n is always 1 . And the solution is pretty obvious :4