Math

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

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asked by Sharee
  1. 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

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    2. 👎 0
    posted by Damon

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