Perhaps the answer is correct, but not the way to get there. Squaring and multiplying with x² on both sides is only possible if you exclude cases such as -1=1 squared -> 1=1.

Another answer will be √i+(1/√i)

Did anyone else do it through imaginary numbers😂 I got 2/sqrt(2) which is the same as sqrt(2) because 2 is just 2^1 and sqrt(2) = 2^0.5 then 2^1/2^0.5 subtraction = 2^0.5 = sqrt(2) so i got the same answer just did not do it his way i started by getting the solutions to the first equation x+1/x = sqrt(2) and multiplied everything by x then got two solutions and through de moivres formula took to power of 777 and could take inverse with de moivre or conjugate of denominator mutliplied by nominator and denominator. I got same answer for both of the solutions of x namely 2/sqrt(2) = sqrt(2)

In fact, x^n + x^-n has a period of 8, when x + 1/x = sqrt(2). And since 777 is one more than a multiple of 8, x^777 + x^-777 = x + x^-1 = sqrt(2). Here is a somewhat tedious proof : For simplicity, note F(n) = x^n + x^-n First, notice that F(n) * F(1) = F(n + 1) + F(n - 1). By decreasing n, we get the following expression : F(n) = F(n-1) * F(1) - F(n-2) F(n) = sqrt(2) * F(n-1) - F(n-2) [1] Then, by subbing [1] into F(n-1), we get : F(n) = 2 F(n-2) - sqrt(2) * F(n-3) - F(n-2) F(n) = F(n-2) - sqrt(2) * F(n-3) Again, subbing [1] into F(n-2), we get : F(n) = sqrt(2) * F(n-3) - F(n-4) - sqrt(2) * F(n-3) F(n) = -F(n-4) [2]. From [2], we get that F(n) = F(n-8).

@ChidexOnlineMathClass01

4 күн бұрын

@@mismis3153 Thanks so much for this

Thank you, very nicely done.

@ChidexOnlineMathClass01

8 күн бұрын

Thanks so much

@ChidexOnlineMathClass01

8 күн бұрын

Thanks so much

I understood

@ChidexOnlineMathClass01

5 күн бұрын

@@gnanadesikansenthilnathan6750 Welcome

hey man the question is wrong as the Range of x+(1/x) is (-∞,-2] U [2,∞) hence 2^1/2 dosent lies between this Range therefore this equation of x+(1/x)=2^1/2 isnt even possible.

@ChidexOnlineMathClass01

6 күн бұрын

How do you think this question can be wrong ?

@ChidexOnlineMathClass01

6 күн бұрын

Obviously the equation does not have real solutions, but that does not make the evaluation of x^777 + 1/x^777 wrong! Infact the question is very correct and very interesting

@Straight_Talk

6 күн бұрын

​​@@ChidexOnlineMathClass01 If there are no real solutions for x, how can there be a real solution for the expression we need to evaluate? Also, it's intuitively wrong to get the same value for the expression when x is raised to the powers of 1/-1 and 777/-777.

@klausolekristiansen2960

6 күн бұрын

@@Straight_Talk In this case, the complex solution is a real number. The real numbers are a subset of the complex numbers.

@Straight_Talk

6 күн бұрын

@@klausolekristiansen2960 What are you talking about? Do you have any grasp of basic maths?