| −b−√Δ | ||
x1= | ||
| 2a |
| 5−3 | 1 | |||
x1= | = | |||
| 2*2 | 2 |
| −b−√Δ | ||
x1= | ||
| 2a |
| 0−30 | −30 | 5 | ||||
x1= | = | =− | ||||
| 2*9 | 18 | 3 |
| 5 | ||
x2= | ||
| 3 |
no to sprawdzam :
a) 4x2−1=(2x−1)(x+3) ⇔ (2x−1)(2x+1)−(2x−1)(x+3)=0 ⇔
⇔ (2x−1)(2x+1−x−3)=0 ⇔ 2(x+12)(x−2)=0 ⇔ x∊{12,2} ,
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
b) 9x2−25=0 ⇔ (3x−5)(3x+5)=0 ⇔ x∊{53−53} ,
TAK
masz rozwiązane bardzo dobrze . ...