Power of reciprocity, reciprocity of power

Since we see powers of $x^{100}={(x^{50})}^2$ and $x^{50}$ , it is reasonable to let $y=x^{50}$ .

Doing so, our original equation becomes $\frac{y^2+1}{y}+\frac{y}{y^2+1}=\frac{101}{10}$ .

Now, let $\frac{y^2+1}{y}=a$ . Then, $\frac{1}{a}=\frac{y}{y^2+1}$ , so we have $a+\frac{1}{a}=\frac{101}{10} \Rightarrow \frac{a^2+1}{a}=\frac{101}{10}$ . Cross-multiplying, we have $10a^2+10=101a \Rightarrow 10a^2-101a+10=0 \Rightarrow (10a-1)(a-10)=0 \Rightarrow a=\frac{1}{10}$ or $a=10$

If $a=10$ , then $\frac{y^2+1}{y}=10 \Rightarrow y^2+1=10y \Rightarrow y^2+10y+1=0 \Rightarrow y=\frac{-10 \pm \sqrt{10^2-4 \cdot 1 \cdot 1}}{2 \cdot 1}=5 \pm 2\sqrt{6}$ .

Taking the 50th roots of these numbers, we have the 4 roots $x= \pm \sqrt[50] {5 \pm 2\sqrt6 }$ . All other roots will be imaginary, so there are 4 roots in this case.

If $a=\frac{1}{10}$ , then $\frac{y^2+1}{y}=\frac{1}{10}$ , and $10y^2-y+1=0$ . Since the discriminant of this, $(-1)^2-4 \cdot 10 \cdot 4=-39<0$ , the equation only has complex roots, and will not produce real roots when taking the 50th root of these roots. This means that there are no roots in this case.

In total, there are $\boxed{4}$ roots.

Let $y = x^{50},$ the equation will be: $\frac{y^2 + 1}{y} + \frac{y}{y^2 + 1} = \frac{101}{10}$

Simplify the equation as we get: $y^2 - 10y + 1 = 0$

Since $y \neq 0,$ we obtain $y = 5 - 2 \sqrt{6}$ or $y = 5 + 2 \sqrt{6}.$

For $y = 5 - 2 \sqrt{6},$ we have $x = \sqrt[50]{5 - 2 \sqrt{6}}$ or $x = -\sqrt[50]{5 - 2 \sqrt{6}}.$

For $y = 5 + 2 \sqrt{6},$ we have $x = \sqrt[50]{5 + 2 \sqrt{6}}$ or $x = -\sqrt[50]{5 + 2 \sqrt{6}}.$

Hence, we get $\boxed{4}$ real values of $x.$

Let $\frac{x^{100}}{1 + x^{50}} = t$

$t + \frac{1}{t} = \frac{101}{10}$

$\Rightarrow 10t^2 - 101t + 10 \Rightarrow t=10,\frac{1}{10}$

Let $x^{50} = y$

Case - 1, $t=\frac{1}{10} \Rightarrow \frac{y^2 + 1}{y} = y + \frac{1}{y} = \frac{1}{10}$ Hence, no real root.

Case-2 $y + \frac{1}{y} = 10 y \Rightarrow y= 5 \pm 2\sqrt{6}$

Hence $x^{50} = 5 \pm 2\sqrt{6} , x =\pm \sqrt[50]{5 \pm 2\sqrt{6}}$ .

Hence , $\fbox{4}$ solutions.

Lets assume $x^{100} + 1 = A$ and $x^{50} = B$ .

Now our equation reduces to:

$\frac{A}{B} + \frac{B}{A} = \frac{101}{10}$

$\Rightarrow \frac{A^2+B^2}{AB} = \frac{101}{10}$

Taking $A^2+B^2=101k$ and $AB = 10k$ . Adding and then subtracting both we get

$(A+B)^2=121k$ and $(A-B)^2=81$ OR $(A+B)^2=121k$ and $(B-A )^2=81$

Eliminating $k$ we get

$\boxed{a=10b}$ OR $\boxed{b=10a}$

*Case 1: * we have $x^{100} + 1 = 10x^{50}$ . Taking $y=x^{50}$

$y^2-10y+1=0$ . Thus we get two values of $y$ . (Both real and positive).

Now, for each value of $y$ which is $x^{50}$ , we get 2 values of $x$ - one positive and one negative. So, total value of $x = \boxed{4}$ /

*Case 2: * we have $10x^{100} + 10 = x^{50}$ . Taking $y=x^{50}$

$10y^2-y+10=0$ . But no real value of $y$ exist in this case.

Therefore total number of possible value of $x = \boxed{4}$

Let's assume $x^{50} = a$ ... Then the equation becomes...

$\Large{\frac{a^2+1}{a} + \frac{a}{a^2+1} = \frac{101}{10}}$

$\Longrightarrow \Large{\frac{a^4+3a^2+1}{a(a^2+1)} = \frac{101}{10}}$

$\Longrightarrow 10a^4-101a^3+30a^2-101a+10 = 0$

Solving for $a$ , we get...

$a = 5 + 2 \sqrt{6}, ~~~5 - 2 \sqrt{6}$

Hence, $x^{50} = 5 + 2 \sqrt{6}, ~~~5 - 2 \sqrt{6}$

$\therefore x = \sqrt[50]{5 + 2 \sqrt{6}},~~~-\sqrt[50]{5 + 2 \sqrt{6}},~~~\sqrt[50]{5 - 2 \sqrt{6}},~~~-\sqrt[50]{5 - 2 \sqrt{6}}$

Total solution: $\large{\fbox{4}}$

Let $y=\dfrac{x^{100}+1}{x^{50}}$ . Then, $y+\dfrac{1}{y}=\dfrac{101}{10}$ $y^2-\dfrac{101}{10}y+1=0$ $y=\dfrac{\dfrac{101}{10}\pm\sqrt{\left(\dfrac{101}{10}\right)^2-4}}{2}$ Now, $\dfrac{x^{100}+1}{x^{50}}=y$ $x^{100}-yx^{50}+1=0$ We can treat this as a quadratic in $x^{50}$ : $x^{50}=\dfrac{y\pm\sqrt{y^2-4}}{2}$ However, we must check whether the values of $y$ are greater than 4. $\left(\dfrac{\dfrac{101}{10}+\sqrt{\left(\dfrac{101}{10}\right)^2-4}}{2}\right)^2>\left(\dfrac{10+9}{2}\right)^2>4$ $\left(\dfrac{\dfrac{101}{10}-\sqrt{\left(\dfrac{101}{10}\right)^2-4}}{2}\right)^2<\left(\dfrac{11-9}{2}\right)^2<4$ Therefore, only the larger value for $y$ works. $x^{50}=\dfrac{y\pm\sqrt{y^2-4}}{2}$ We can choose either the plus or the minus value, and $x$ can be positive or negative, so there are $2\cdot 2=\boxed{4}$ real values for $x$ .

$x^{50}+1/x^{50}=a$ then /(a+1/a = 101/10) hence $a= 10$ so $x^{50}+1/x^{50}= 10$ now if $x^{50}=p$ then its a quadratic equation there are 2 +ve solution hence "x" will have 4 solution

We will solve this by substitution.

Let y = x^50. Then the equation becomes: (y^2+1)/y + y/(y^2+1) = 101/10 Cross multiply: (y^2+1)^2 + y^2 = 101/10(y^2+1)y Expand and simplify: 10y^4 + 30y^2 + 10 = 101y + 101y^3 Rearrange to get: 10y^4 - 101y^3 + 30y^2 - 101y + 10 = 0. Since y=0 is not a solution, we can divide by y^2 to get: 10y^2 - 101y + 30 - 101y^-1 + 10y^-2 = 0. Factorise: 10(y+1/y)^2 - 101(y+1/y) + 10 = 0.

Now substitute again, let z = y+1/y. Equation is now: 10z^2 - 101z + 10 = 0. z = 10 or 1/10. Reject 1/10 as there is no real value of y for which y+1/y = 1/10. Hence z = 10 --> y + 1/y = 10. Both of the solutions of y are positive real numbers (a) and (b) For x^50 = a, there are 2 solutions as x can be positive or negative. Likewise for x^50 = b.

Therefore 4 real values of x satisfy this equation

Let y = x^50+1/x^50. So

y+1/y = 101/10,

10y^2-101y+10 = 0,

(y-10)(10y-1) = 0,

x^50+1/x^50 = y = 10 or 1/10,

If y = 10, then x^100-10x^50+1 = 0, whose roots are x^50 = 5+/-2 * sqrt(6), both being positive. Thus there are 4 real roots for y = 10, namely x = +/-[5+/-2*sqrt(6)]^(1/50).

If y = 1/10, then 10x^100-x^50+10 = 0, whose discriminant is -399, or whose roots are complex. Thus there are 0 real root for y = 1/10.

In all, there 4 real values of x satisfy the given equation.

Let $a = x^{50}$ and $c = \frac{a^{2}+1}{a}$ . So $\frac{a^{2}+1}{a}\ + \frac{a}{a^{2}+1} = \frac{101}{10}$ thus $c + \frac{1}{c} = \frac{101}{10}$ .

After multiplying by $10c$ and rearranging, we obtain the quadratic equation $10c^{2} - 101c + 10 = 0$ . After factorisation into $(10c-1)(c-10) = 0$ , solving it yields $c = 10, \frac{1}{10}$ .

So $\frac{a^{2}+1}{a} = 10$ or $\frac{a^{2}+1}{a} = \frac{1}{10}$ .

For $\frac{a^{2}+1}{a} = \frac{1}{10}$ , multiplying by $10a$ and rearranging yields $10a^{2}-a+10 = 0$ . Consider discriminant which equals $(-1)^{2} - 4(10)(10) < 0$ .

This shows there are no real roots $a$ of the equation and consequently no corresponding $x$ .

For $\frac{a^{2}+1}{a} = 10$ , multiplying by $a$ and rearranging gives $a^{2}-10a+1 = 0$ . By the quadratic formula, we get $a = 5 + 2 \sqrt{6}, 5 - 2 \sqrt{6}$ .

As $5 > 2 \sqrt{6} > 0$ , we have solutions $a_{1},a_{2} > 0$ .

As $a = x^{50}$ , we have $x = \pm \sqrt[50]a_{1}, \pm \sqrt[50]a_{2}$ , all of which are clearly distinct.

Therefore there are 4 real solutions $x$ satisfying the equation above.

Let's define $a = (x^{100} + 1)$ , $b = x^{50}$ : we can therefore rewrite the expression as

$\frac{a}{b} + \frac{b}{a} = \frac{101}{10} \iff a^2 + \frac{101 b}{10}a + b^2 = 0$

Solving for $a$ we have 2 roots $a_1= \frac{b}{10}, a_2 = 10b$ so we can factorize as:

$(a - 10b) (a - \frac{b}{10})$

Defining $y = x^{50}$ we have $a = y^2 +1$ , $b = y$ and we can rewrite as:

$(y^2 -10y +1) (y^2 - \frac{y}{10} + 1)$

• $y^2 - \frac{y}{10} + 1$ has $\Delta = \frac{1}{100} - 4 < 0$ so can't have real roots.

• $y^2 -10y +1$ has 2 real roots $y_{1,2} = 10 \pm \sqrt{96}$

Since $y_1 >0, y_2 > 0$ we have 4 real solutions:

$x_{1,2} = \pm \sqrt[50]{y_1}$ , $x_{3,4} = \pm \sqrt[50]{y_2}$

1. common denominator is $x^{150} + x^{50}$
2. then the LHS becomes $\frac{x^{200} + 3x^{100} + 1} {x^{150} + x^{50}}$
3. multiplying both sides by $x^{150} + x^{50}$ gives $x^{200} + 3x^{100} + 1 = 10.1x^{150} + 10.1x^{50} = 0$
4. multiplying by $10$ and rearranging gives $10x^{200} - 101x^{150} + 30x^{100} -101x^{50} + 10 = 0$
5. factor this to get $(10x^{100} - x^{50} + 10) (x^{100} - 10x^{50} + 1)$
6. $(10x^{100} - x^{50} + 10)$ has no real solutions because $\triangle = -399$ , but $(x^{100} - 10x^{50} + 1)$ gives solutions of $5 + 2\sqrt{6}$ and $5 - 2\sqrt{6}$ for $x^{50}$
7. then the solutions are $x = \pm \sqrt [50]{5 + 2\sqrt{6}}$ and $x = \pm \sqrt [50]{5 - 2\sqrt{6}}$
8. so there are $\boxed{4}$ real solutions for $x$