Wyznacz współczynniki b i c trójmianu kwadratowego y=x^2+bx+c którego pierwiastkami są:
a) 2,5
b) -3 -9
c) 1/2, 4
d) -1/3, 1/4
e) -2pierw z 2 , pierw 2
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a)
0=2^2+2b+c
0=5^2+5b+c
0=4+2b+c
0=25+5b+c
c=-4-2b
0=25+5b-4-2b
c= -4-2b
0=21+3b
-21=3b
b=-7
c= -4 -2*(-7)
c= -4 + 14
c = 10
b= -7
b)
0 = (-3)^2 -3b +c
0 = (-9)^2 - 9b +c
0 = 9 - 3b +c
0 = 81 - 9b +c
c= -9 +3b
0= 81 - 9b - 9 + 3b
0= 72 - 6b
6b = 72
b = 12
c = -9 + 3*12
c = 27
c)
1/2, 4
0=1/2^2 +1/2b +c
0=16+4b+c
0=1/4+1/2b+c
0=16+4b+c
c= -1/4 - 1/2b
0= 16 + 4b -1/4 - 1/2b /*2
0=32 + 8b - 1/2 - b
0=31 i 1/2 + 7b
-31 i 1/2 = 7b
b=-4,5
c= - 1/4 - 1/2 * (-4,5)
c = -0,25 + 6,75
c = 6,5
d) -1/3, 1/4
0=(-1/3)^2 -1/3b +c
0=(1/4)^2 +1/4b +c
0=1/9-1/3b+c
0=1/16 + 1/4b +c
c= -1/9 + 1/3b
0 = 1/16 + 1/4b - 1/9 + 1/3b
Och, nienawidzę ułamków i pierwiastków ;D
0= 9/144 + 3/12b - 16/144 + 4/12b
-7/144 + 7/12b = 0
7/12b = 7/144
7b = 7/144 * 12
7b = 7/12
b = 1/12
c = -1/9 + 1/3 * 1/12
c = -1/9 + 1/36
c = -4/36 + 1/36
c = -3/36
c = -1/12
e) -2pierw z 2 , pierw 2
0=(-2√2)^2 - 2√2b +c
0 = √2^2 +√2b +c
0= 4*2 - 2√2b + c
0 = 2 +√2 +c
0= 8 -2√2 +c
0 = 2+√2b +c
c = -8 + 2√2b
0 = 2 +√2b - 8 +2√2
0 = -6 + 3√2b
3√2b = 6 /:3
√2b = 2
b = 2 / √2
b = 2 √2 / 2
b = √2
c = -8 + 2 √ 2 * √2
c = -8 + 2*2
c = -8 + 4
c = -4
b = √ 2
c = -4
W(x)=x²+bx+c
a)
W(2)=0
W(5)=0
0=5²+5b+c
0=2²+2b+c
0=25+5b+c
0=4+2b+c
-5b-c=25
2b+c=-4
----------+
-3b=21
b=-7
b=-7
c=10
b)
W(-3)=0
W(-9)=0
0=(-3)²-3b+c
0=(-9)²-9b+c
0=9-3b+c
0=81-9b+c
-3b+c=-9
9b-c=81
---------+
6b=72
b=12
b=12
c=27
c)
W(0,5)=0
W(4)=0
0=0,5²+0,5b+c
0=4²+4b+c
0=0,25+0,5b+c
0=16+4b+c
0,5b+c=-0,25
-4b-c=16
------------+
-3,5b=15,75
b=-4,5
b=-4,5
c=2
d)
W(-¹/₃)=0
W(0,25)=0
0=0,25²+0,25b+c
0=(-¹/₃)²-¹/₃b+c
0=0,0625+0,25b+c
0=¹/₉-¹/₃b+c
-0,25b-c=0,0625
-¹/₃b+c=-¹/₉
----------------+
-⁷/₁₂b=-⁷/₁₄₄
-84b=-7
84b=7
b=¹/₁₂
b=¹/₁₂
c=-¹/₁₂
e)
W(-2√2)=0
W(√2)=0
0=(√2)²+b√2+c
0=(-2√2)²-2b√2+c
0=2+b√2+c
0=8-2b√2+c
-2=b√2+c
8=2b√2-c
-----------+
3b√2=6
b√2=2
b=√2
b=√2
c=-4