Homework Help: Math

Posted by Sharee on Thursday, August 5, 2010 at 2:06pm.

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

Math - Damon, Thursday, August 5, 2010 at 3:07pm
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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