x1 | x2 | |||
Ulozyc nowe równanie kwadratowe którego pierwiastki bylyby rowne | i | |||
x2 | x1 |
x1 | x2 | |||
Jeśli sobie oznacze t1= | i t2= | to | ||
x2 | x1 |
x1 | x2 | x12+x22 | x1+x2)2−2x*x2 | |||||
t1+t2= | + | = | = | |||||
x2 | x1 | x1*x2 | x1x2 |
−b | c | |||
x1+x2= | x1*x2= | |||
a | a |
(−b/a)2−2*(c/a) | (b2/a2)−2(c/a) | b2−2ac | ||||
Wiec | = | = | ||||
c/a | c/a | ac |
x1 | x2 | x1*x2 | ||||
t1*t2= | * | = | = 1 | |||
x2 | x1 | x1*x2 |
b2−2ac | ||
t2− | t+1=0 | |
ac |
x1 | x2 | |||
(x− | )(x− | )=0 | ||
x2 | x1 |
x1 | x2 | |||
x2− | x− | x+1=0 | ||
x2 | x1 |
x12+x22 | ||
x2−x( | )+1=0 | |
x1x2 |
(x1+x2)2 | ||
x2−x( | −2)+1=0 | |
x1x2 |
| ||||||||
x2−x( | −2)+1=0 | |||||||
|
b2 | ||
x2−x( | −2)+1=0 | |
ac |