Because we are considering only the principal square root which is the positive side of the root, disregarding of what's inside. That's explaining the left hand side of the equation. As for the denominator you evaluate what's the minimum "x" can be without turning numerator or denominator negative which is 1/7, 3 or 4, so if you go 1/7 to infinity the numerator will always be positive, that leaves us with the denominator that must always be positive as well or you end up with an imaginary number.
I have never seen the “number line method” when you checked the intervals before! Could you explain how you knew that x>0 in the very beginning?
Because we are considering only the principal square root which is the positive side of the root, disregarding of what's inside. That's explaining the left hand side of the equation. As for the denominator you evaluate what's the minimum "x" can be without turning numerator or denominator negative which is 1/7, 3 or 4, so if you go 1/7 to infinity the numerator will always be positive, that leaves us with the denominator that must always be positive as well or you end up with an imaginary number.
The given equation is written as x^4-7x^3+12x^2-7x+1=0 > (x+1/x)^2 -t(x+1/x_ + 10 =0 > x+1/x = 2,5 > x = 1, 1/2[5 +/-√21]
(x-1)(x-1)(x^2-5x+1)=0 , x= 1 , (5+V21)/2 , (5-V21)/2 ,
Даже придираться не к чему