Donna P. answered 12/06/15
Tutor
New to Wyzant
Algebra I, Algebra II and Geometry tutor
To find the number of solutions you would use the discriminant. This is the part of the quadratic formula under the radical.
b2 - 4ac
the a,b, and c come from the coefficients in the quadratic formula ax2 + bx + c = 0
To find number of real solutions you calculate the value of the discriminant
If the value is greater than 0 then there are 2 real solutions
If the value is equal to 0 then there is 1 real solution
If the value is less than 0 then there are 0 real solutions (here the solutions would be imaginary)
For your problem x^2+6x+c=0 a = 1, b = 6 and c = c
So b2 - 4ac = 62 - 4(1)c = 0 since we need one solution
36 - 4c = 0
+4c +4c
36 = 4c
36/4 = 4c/4
9 = c
CHECK
x^2+6x+c=0
x2 + 6x + 9 = 0
(x + 3)(x + 3) = 0
x + 3 = 0
X = -3 This is the only solution.