How do you solve #9x ^ { 2} - 36x =6#?

2 Answers
Jun 12, 2017

Answer:

#x =(36 +- 6sqrt42) / (18) #

Explanation:

#9x^2 - 36x = 6#

We can move #6# to the other side, then use the quadratic formula.

#9x^2 - 36x -6 = 0#

#ax^2 + bx +c = 0#

#a = 9#
#b = -36#
#c = -6#

# x = (-b +- sqrt(b^2-4ac)) / (2a) #

# x = (- -36 +- sqrt((-36)^2-4 xx 9 xx -6)) / (2 xx 9) #

Now we can simplify and solve.

# x = (cancel(- -)36 +- sqrt(-36^2-4 xx 9 xx -6)) / (18) #

# x =(36 +- sqrt((-36)^2- 36 xx -6)) / (18) #

# x =(36 +- sqrt((-36)^2 + 216)) / (18) #

# x =(36 +- sqrt(1296 + 216)) / (18) #

# x =(36 +- sqrt(1512)) / (18) #

#color(blue)( x =(36 +- 6sqrt42) / (18) #

#therefore # #x =(36 +- sqrt(1512)) / (18) #

Jun 12, 2017

Answer:

#x = (6 +- sqrt42)/3#

Explanation:

#y = 9x^2 - 36x - 6 = 0#
#y = 3(3x^2 - 12x - 2) = 0#
Use the improved quadratic formula (Google, Socratic Search):
#D = d^2 = b^2 - 4ac = 144 + 24 = 168# --> #d = +- 2sqrt42#
There are 2 real roots:
#x = -b/(2a) +- d/(2a) = 12/6 +- (2sqrt42)/6 = 2 +- sqrt42/3 =#
#x = (6 +- sqrt42)/3#