Algebra one, semester one, unit five, assignment: the
discriminant
For each problem I found out what a, b, and c were. I filled in the
format with the numbers that equaled a, b and c. I multiplied the
exponent first, then four times a and c. The final thin I did was
subtract the numbers and write how many real number solutions
there were for the problem.
B^2-4*a*c
X^2+6x-3=0
A=1 b=6 c=-3
6^2-4*1*-3
36-4*1*-3
48
There are two real number solutions
3x^2+2x+1=0
A=3 b=2 c=1
2^2-4*3*1
4-4*3*1
-8
There are no real number solutions
X^2+4x+4=0
A=1 b=4 c=4
4^2-4*1*4
16-4*1*4
0
There is one real number solution
5x^2+x=4