TEKST ZADATKA
Za koje vrednosti realnih parametra m i k rešenja x1 i x2 jednačina zadovoljavaju date relacije (zadaci 312-315)? x2+(2m+2)x+m=0, x121+x221>8.
REŠENJE ZADATKA
Da bi rešenja kvadratne jednačine bila realna, diskriminanta mora biti veća ili jednaka nuli (D≥0). Računamo diskriminantu:
D=(2m+2)2−4⋅1⋅m=4m2+8m+4−4m=4m2+4m+4=4(m2+m+1) Kvadratni trinom m2+m+1 je uvek pozitivan jer je njegova diskriminanta negativna (12−4⋅1⋅1<0), pa je D>0 za svako realno m. Rešenja su uvek realna.
Primenjujemo Vijetove formule za datu kvadratnu jednačinu:
x1+x2x1⋅x2=−ab=−(2m+2)=ac=m Transformišemo zadatu nejednačinu svodeći je na zajednički imenilac, kako bismo mogli da upotrebimo Vijetove formule:
x121+x221=(x1⋅x2)2x12+x22>8 Izražavamo zbir kvadrata rešenja x12+x22 preko zbira i proizvoda rešenja:
x12+x22=(x1+x2)2−2x1x2 Zamenjujemo vrednosti iz Vijetovih formula u dobijeni izraz:
x12+x22=(−(2m+2))2−2m=4m2+8m+4−2m=4m2+6m+4 Zamenjujemo dobijeni izraz i proizvod rešenja u transformisanu nejednačinu:
m24m2+6m+4>8 Prebacujemo sve na levu stranu i svodimo na zajednički imenilac:
m24m2+6m+4−8m2>0⟹m2−4m2+6m+4>0 Pošto se parametar m nalazi u imeniocu, mora važiti uslov da je imenilac različit od nule:
m2=0⟹m=0 Za m=0, imenilac m2 je uvek strogo pozitivan. Da bi ceo razlomak bio pozitivan, brojilac mora biti pozitivan:
−4m2+6m+4>0 Delimo nejednačinu sa −2 (pri čemu se znak nejednakosti menja):
2m2−3m−2<0 Rešavamo odgovarajuću kvadratnu jednačinu 2m2−3m−2=0 da bismo našli nule trinoma:
m1,2=2⋅2−(−3)±(−3)2−4⋅2⋅(−2)=43±9+16=43±5 Nule kvadratnog trinoma su:
m1=48=2,m2=4−2=−21 Faktorišemo kvadratni trinom kako bismo odredili znak pomoću tabele:
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m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) m∈(−∞,−1/2) m∈(−1/2,2) m∈(2,+∞) 2(m−2)(m+1/2) Na osnovu tabele, kvadratni trinom je negativan na intervalu:
m∈(−21,2) Uzimajući u obzir početni uslov za imenilac m=0, konačno rešenje za parametar m je unija intervala:
m∈(−21,0)∪(0,2)