Janek191:
cd.
x
2 + 4 x − 21 = 0
Δ = b
2 − 4ac = 16 − 4*1*(−21) = 16 + 84 = 100
√Δ = 10
| | − 4 − 10 | | − 4 + 10 | |
x = |
| = − 7 lub x = |
| = 3 |
| | 2 | | 2 | |
zatem x
2 + 4 x − 21 = ( x + 7)*( x − 3)
Mamy więc
( x − 4)*[ ( x + 4)*( x + 2) − ( x + 7)*( x − 3)] = 0
( x − 4)*[ x
2 + 2 x + 4x + 8 − ( x
2 − 3x + 7 x − 21)] = 0
(x −4)*[ 6x + 8 − 4x +21] = 0
( x − 4)*( 2x + 29) = 0
x − 4 = 0 lub 2x = − 29
| | 29 | |
x = 4 lub x = − |
| = − 14,5 |
| | 2 | |
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