Asked by Sharee
Use the discriminant to determine how many real-number solutions the equation has.
Problem 1
w2 - 2w + 2 = 0
A) 2
B) 1
C) 0
Problem 2
1 - 7a2 = -7a - 2
A) 2
B) 1
C) 0
Problem 1
w2 - 2w + 2 = 0
A) 2
B) 1
C) 0
Problem 2
1 - 7a2 = -7a - 2
A) 2
B) 1
C) 0
Answers
Answered by
Damon
w2 - 2w + 2 = 0
a = 1
b = -2
c = 2
b^2-4ac = 4-4(1)(2) = 4-8 = -4
sqrt (-4) is imaginary
no real solutions (although 2 complex solutions)
1 - 7a2 = -7a - 2
7a^2 -7a -3 = 0
b^2-4ac = 49 -4(7)(-3) = 133
sqrt 133 is real number >0
so two real solutions
a = 1
b = -2
c = 2
b^2-4ac = 4-4(1)(2) = 4-8 = -4
sqrt (-4) is imaginary
no real solutions (although 2 complex solutions)
1 - 7a2 = -7a - 2
7a^2 -7a -3 = 0
b^2-4ac = 49 -4(7)(-3) = 133
sqrt 133 is real number >0
so two real solutions
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