[if we write x, as sqrt(6+sqrt(6+x)) with in the left side of the given equation again and again .. we get, sqrt(6+sqrt(6+sqrt(6.....)))) would ve the solution of the same question.. so; sqrt(6+A)=A would be the same question. so the solution would be the pozitive root of, A^2-A-6=0 which is 3.] may be thought as a simple additional method. thanks for sharing.
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[if we write x, as sqrt(6+sqrt(6+x)) with in the left side of the given equation again and again .. we get, sqrt(6+sqrt(6+sqrt(6.....)))) would ve the solution of the same question.. so; sqrt(6+A)=A would be the same question. so the solution would be the pozitive root of, A^2-A-6=0 which is 3.] may be thought as a simple additional method. thanks for sharing.
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x^4-12x^2-x+30=0 , (x-3)(x^3+3x^2--3x-10)=0 , x=3 , (x+2)(x^2+x-5)=0 , result , x= 3 , -2 , (-1+V(21))/2 , (-1-V(21))/2 ,
solu , x=3 ,
x>0
x=3, or x=(-1+rt21)/2
Two solutions.
@@에스피-z2g I checked that (-1+V(21))/2=1.79129 is not a solution. Substituting it, we get ~2.96501. The solution is x=3.
x=3, by inspection.
@@prollysine Yes,it is.
Thanks for your kind
explanation.
@@skrachit solu , x=3 ,
For real solutions, √(6 + x) ≥ 0 , x ≥ 0 , x = √(6 + √(6 + x)) ≥ √6 => x ≥ √(6 + √6)
let y = √(6 + x) ≥ 0, 6 + x = y^2 & 6 + y = x^2
subtracting, x - y = y^2 - x^2 => (y - x)(y + x + 1) = 0
since y + x + 1 ≠ 0, y - x = 0 => 6 + x = x^2 => x^2 - x - 6 = (x - 3)(x + 2) = 0 => x = 3