Wow! Very well done. I followed this problem perfectly from beginning to end. They surely take a lot to write out and go thru, so worth it to show all the steps and explain the dteps as you go. Your work is something to be proud of Its math instructors ss yourself who i admire!
warum so kompliziert ? auf beiden Seiten ist die gleiche gerade Potenz, also muss sein +/- x = +/- (x-1). Das sins 4 mögliche Lösungen, von denen 2 richtig sindmit x = 1/2
This is literally mumbo-jumbo, complete nonsense!! There is no number for x that satisfies the equation. Try this: x = x - 1. There is no value of x such that subtracting off one gives the same number. x can't equal itself minus 1. Jibberish, jibberish, jibberish! Subtract 1 from both side. You get 0 = -1 !!!!! He says one solution is 1/2. Go stick 1/2 back into the original equation, the two sides are not equal.
x=1/2 is a double root. In any polynomial with real coefficients, all the roots are either real or complex conjugates. There are two pairs of complex conjugate numbers, and fifth real root, therefore the sixth one has to be real as well, but in this particular case, all the solutions must have a real part of 1/2, therefore the 1/2 itself must be a double root.
@@Vojtaniz01actually, this is not a 6th degree polynomial, but a 5th degree polynomial. If you write it in standard form you get 0x^6-6x^5+15x^4-20x^3+15x^2-6x+1... but the leading coefficient of a polynomial cannot be zero, so the highest degree of this polynomial must be 5 instead of 6. He showed all 5 solutions and there is not a double root. Good intuition though, I thought the same at first.
