find the value of C

Question

Find the value of c that makes the equation x^2+6x+c=0 have one real solution 

Answer 1

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.