How do you solve #x^2 + 9x + 9 = 0# using the quadratic formula?

2 Answers
Apr 26, 2016

Answer:

Root1=# -1.146# and Root2 = # -7.854#

Explanation:

This is quadratic equation where # a=1 ;b=9; c=9; b^2-4ac > 0# So roots are real. Roots are # -b/(2*a) +- sqrt (b^2-4ac)/(2a)# #:.#Root1 #=-9/2+sqrt45/2 = -1.146# and Root2#=-9/2-sqrt45/2 = -7.854#[Ans]

Apr 26, 2016

Answer:

#(-9 +- 3sqrt5)/2#

Explanation:

#y = x^2 + 9x + 9 = 0#
Use the improved quadratic formula (Socratic Search)
#D = d^2 = b^2 - 4ac = 81 - 36 = 45 #--> #d = +- 3sqrt5#
There are 2 real roots:

#x = -b/(2a) +- d/(2a) = -9/2 +- 3sqrt5/2 = (-9 +- 3sqrt5)/2#