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2014-05-06T22:03:18+03:00
Ax²+(3a+1)x+a+3=0
Δ=(3a+1)²-4a(a+3)
Δ=9a²+6a+1-4a²-12a
5a²-6a+1=0
Δ=36-20=16
a1=(6-4)/10=2/10 simplificam=1/5
a2=(6+4)/10=10/10=1
1) daca a=1/5 atunci ecuatia devine:1/5·x²+(3/5+1)x+1/5+3=0 aducem la acelasi numitor⇒
x²+8x+16=0
X1=X2=-8/2=-4
2) daca a=1 atunci ec.devine:x²+4x+4=0
X1=X2=-4/2=-2
2 5 2