x2 - 6x + 1 = 0
x = [-(-6) +&- square root of [(-6)2 - 4(1)(1)]]/2(1)
x = (6 +&- square root of 32)/2
x = [6 +&- 4(square root of 2)]/2
x = 3 +&- 2(square root of 2)
x = 3 + 2(square root of 2) or x = 3 - 2(square root of 2)
Check:
Yes because if 1+0=1 than 0 plus b equals b
1 plus 1 plus 1 plus 1 equals 1 times 4. 1 times 4 equals 4. 4 minus 4 equals 0. 0
If it's 1 +-1 =0
0 because X= -1
(11, 1, 0) and (0, 1, 11) are the two possible solutions.
Yes because if 1+0=1 than 0 plus b equals b
1 plus 1 plus 1 plus 1 equals 1 times 4. 1 times 4 equals 4. 4 minus 4 equals 0. 0
It equals 7
It equals 6
If it's 1 +-1 =0
x = 0
0 because X= -1
because when you add 1 to 0 it is 2
(11, 1, 0) and (0, 1, 11) are the two possible solutions.
Anything multiplied by 0 is 0, so all the ones before the 0 can be ignored. It is now 0+1 which equals 1.
Not in normal arithmetic.
1