Posted by Sharee on .
Use the discriminant to determine how many realnumber 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,
w2  2w + 2 = 0
a = 1
b = 2
c = 2
b^24ac = 44(1)(2) = 48 = 4
sqrt (4) is imaginary
no real solutions (although 2 complex solutions)
1  7a2 = 7a  2
7a^2 7a 3 = 0
b^24ac = 49 4(7)(3) = 133
sqrt 133 is real number >0
so two real solutions