Adamm: (x+2)
2−1+(x+3)
3+(x+4)
4−1=0
(x+1)(x+3)+(x+3)
3+[(x+4)
2+1](x+3)(x+5)=0
x+3=0 lub x+1+(x+3)
2+[(x+4)
2+1](x+5)=0
x+3=0 lub x+5+(x+3)
2−4+[(x+4)
2+1](x+5)=0
x+3=0 lub x+5+(x+1)(x+5)+[(x+4)
2+1](x+5)=0
x+3=0 lub x+5=0 lub (x+4)
2+x+3=0
x+3=0 lub x+5=0 lub x
2+9x+19=0
Δ=5
| | −9+√5 | | −9−√5 | |
x=−3 lub x=−5 lub x= |
| lub x= |
| |
| | 2 | | 2 | |
| | −9+√5 | | −9−√5 | |
x∊{−3;−5; |
| ; |
| } |
| | 2 | | 2 | |
Mila:
II sposób
x+4=t
t
4+(t−1)
3+(t−2)
2−2=0⇔
t
4+t
3−2t
2−t+1=0
W(1)=0
Schemat Hornera
1 1 −2 −1 1 t=1
1 2 0 −1 0
−−−−−−−−−−−−−−
t
4+t
3−2t
2−t+1=(t−1)*(t
3+2t
2−1)
P(t)=(t
3+2t
2−1)
P(−1)=−1+2−1=0
Schemat Hornera: t=−1
1 2 0 −1
1 1 −1 0
−−−−−−−−−−−−−−−−
t
4+t
3−2t
2−t+1=(t−1)*(t+1)*(t
2+t−1)
(t
2+t−1)=0
Δ=5
| | −1−√5 | | −1+√5 | |
t= |
| lub t= |
| |
| | 2 | | 2 | |
========================
| | −1−√5 | | −1+√5 | |
x+4=1 lub x+4=−1 lub x+4 = |
| lub x+4= |
| |
| | 2 | | 2 | |
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