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〖(2+√3)〗^(x^2-2x+1)+〖(2-√3)〗^(x^2-2x-1)=2/(2-√3)
The above equation can be rewritten as
〖(2+√3)〗^(x^2-2x)*(2+√3)+〖(2-√3)〗^(x^2-2x)*(2-√3)-1 = 2/(2-√3)
Multiply the equation with 2-√3
We get〖(2+√3)〗^(x^2-2x)*(2+√3)(2-√3)+〖(2-√3)〗^(x^2-2x)*(2-√3)-1(2-√3)=4
Or 〖(2+√3)〗^(x^2-2x)*(4-3)+ 〖(2-√3)〗^(x^2-2x)*(2-√3)-1+1=4 ( Using (a+b)*(a-b)=a^2-b^2 in the first part on left side and law of indices in second part a^m*a^n=a^(m+n) and remembering that a^0=1)
〖(2+√3)〗^(x^2-2x)+〖(2-√3)〗^(x^2-2x)=2
In second part on left side we can multiply numerator and denominator with conjugate 2+√3
The equation now becomes
〖(2+√3)〗^(x^2-2x)+〖[(2-√3)(2+√3)/(2+√(3)])〗^(x^2-2x)=2
We again use (a+b)*(a-b)=a^2-b^2 in the second part on LHS (2-√3)(2+√3)=4-3=1
The equation now can be re-written as
〖(2+√3)〗^(x^2-2x)+〖[1/(2+√(3)])〗^(x^2-2x)=2
Let (2+√3)^(x^2-2x)=y
The above equation is now
Y+1/y = 2
Or (y²+1)/y =2
Arranging the above equation we get
Y²-2Y +1 = 0
hope you can do iy now
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