| 1 | 1 | |||
( | )x+2 + ( | )x−42=0 | ||
| 4 | 2 |
| 1 | 1 | |||
(( | )2)x+2 +( | )x −42=0 | ||
| 2 | 2 |
| 1 | 1 | 1 | 1 | 1 | |||||
x+2= | * | x= | * | 2x
| |||||
| 4 | 16 | 4 | 16 | 2 |
czyli x=log12 6
| 1 | |
t2 + t − 42 = 0 | |
| 16 |
| 1 | ||
Δ = 1 + 10 | ||
| 2 |
| √46 | ||
√Δ = | t > 0 | |
| 2 |
| √46 | ||
t1 = 4 * (−1 − | ) < 0 | |
| 2 |
| √46 | ||
t2 = 4 * (−1 + | ) = −4 + 2√46 | |
| 2 |
| 1 | ||
( | )x = −4 + 2√46 | |
| 2 |
| √46 | ||
t1 = 8 * (−1 − | ) < 0 | |
| 2 |
| √46 | ||
t2 = 8 * (−1 + | ) = −8 + 4√46 | |
| 2 |