Asked by guy person
How many real number solutions does this equation have?
0= 5x^2 +2x - 12
one, two none, or infinite
How many real number solutions does this equation have?
0= 2x^3 - 20x + 50
0= 5x^2 +2x - 12
one, two none, or infinite
How many real number solutions does this equation have?
0= 2x^3 - 20x + 50
Answers
Answered by
Amy
If the discriminant is negative no real solution exists.
If the discriminant is equal to 0 only one real solution exists.
If the discriminant is positive 2 real solutions exist.
The discriminant:
D = b² - 4 ac
0= 5x^2 +2x - 12
here a = 5, b = 2, c = -12
D = b² - 4 ac
D = (2)^2 - 4 (5)(-12) = 244
Since the discriminant is positive, 2 real solutions exist.
0= 2x^3 - 20x + 50
here a=2, b=-20, c=50
D = b² - 4 ac
D =(-20)^2 - 4(2)(50) = 0
Since the discriminant is equal to 0 only one real solution exists.
If the discriminant is equal to 0 only one real solution exists.
If the discriminant is positive 2 real solutions exist.
The discriminant:
D = b² - 4 ac
0= 5x^2 +2x - 12
here a = 5, b = 2, c = -12
D = b² - 4 ac
D = (2)^2 - 4 (5)(-12) = 244
Since the discriminant is positive, 2 real solutions exist.
0= 2x^3 - 20x + 50
here a=2, b=-20, c=50
D = b² - 4 ac
D =(-20)^2 - 4(2)(50) = 0
Since the discriminant is equal to 0 only one real solution exists.
Answered by
guy person
oh thanks, that little guide at the top makes a lot more sense than the way my teacher explained it. THANKS!!!
Answered by
henry2,
1. 5x^2+2x-12 = 0.
h = Xv = -B/2A = -2/10 = -0.2.
K = 5*(-0.2)^2 + 2*(-0.2)-12 = -12.2.
V(h, k) = V(-0.2, -12.2).
k < 0. Therefore, we have 2 solutions.
k = 0. One solution.
k > 0. No solutions.
h = Xv = -B/2A = -2/10 = -0.2.
K = 5*(-0.2)^2 + 2*(-0.2)-12 = -12.2.
V(h, k) = V(-0.2, -12.2).
k < 0. Therefore, we have 2 solutions.
k = 0. One solution.
k > 0. No solutions.
Answered by
Billy Haybottom
formula: -b±√b^2 -4ac
____________
2a
____________
2a
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