To figure this out, you need to find the discriminant which is
(b^2 - 4ac). If you're discriminant is negative, then you have imaginary roots.
A) (-1)^2 - (4)(1)(-1) = 5
B) (0)^2 - (4)(1)(-2) = 8
C) (1)^2 - (4)(1)(1) = -3
D) (0)^2 - (4)(1)(-4) = 16
C has imaginary roots.
2007-11-16 12:09:16
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answer #1
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answered by npontello12 2
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C) x^2+x+1=0
2007-11-16 12:13:36
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answer #2
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answered by seeleeree 3
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A)x^2-x-1=0
Discriminant = 1+4 =5>0 , real solutions only
B)x^2-2=0
Discrimianant = 0 +4(2) =8 >0 , real solutions only
C)x^2+x+1=0
Discriminant = 1 -4 = -3 <0 only complex solutions
x1 = (-1 - i sqrt(3) )/2
x2 = (-1 + i sqrt(3) )/2
D)x^2-1=0
(x -1)( x+1)=0 : solutions are x = 1 and x =-1
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Summary
C) has has complex roots
2007-11-16 12:09:03
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answer #3
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answered by Anonymous
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1x1-4(1)(1)= -3 <0
x^2+x+1=0
2007-11-16 12:15:03
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answer #4
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answered by Lumi 2
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2016-12-16 10:57:50
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answer #5
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answered by mcintire 4
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find middle^@-4Front*Back
aka b^2-4ac
If Pos, both are real
if 0, just one real root, no im roots
if neg, two im roots
A)the value is 5
B)the value is 4
C)the value is -3
D)the value is 4
So, C is the one with imaginary roots.
Or, you could graph it. THe one that doesn't intersect the x axis has imaginary roots.
2007-11-16 12:09:53
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answer #6
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answered by SaintPretz59 4
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C). Calculate the discriminant for each of these.
The equation has imaginary roots if and only if
the discriminant, b²-4ac of the equation ax²+bx+c is
negative.
The discriminants are:
A). 5
B). 8
C). -3
D). 4
The only one of these which is negative is C),
so this is the only equation with imaginary roots.
2007-11-16 14:58:51
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answer #7
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answered by steiner1745 7
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