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zbiorem rozwiazan nierownosci x²+bx+c<0 jest zbior A={x∈R; x∈(0,4)}. wyznacz wartosci wspolczynnikkow b i c.
bardzo dokladne obliczenia
odp: b=-4, c=0
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a>0
Xw=2 (os symetrii paraboli x=2)
-b/2a=2
-b/2=2 /*2
-b=4
b=-4
oraz ze wzorow Viete'a
x1*x2= c/a
c=0*4=0
Odp. b=-4, c=0
x^2 + bx + c < 0
Mamy
x1 = 0, x2 = 4
oraz a = 1 > 0
zatem
p = (x1 +x2)/2 = 4/2 = 2
p = - b/(2a)
2 = -b/2 => b = - 4
b = - 4
============
x1*x2 = c/a
0*4 = c /1
0 = c/1
c = 0
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Odp. b = - 4; c = 0
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