How do you solve # 2x^2 - 12x + 7 = 5#?

2 Answers
Mar 28, 2018

Answer:

#x = 3+-2sqrt(2)#

Explanation:

Given:

#2x^2-12x+7=5#

Subtract #5# from both sides to get:

#2x^2-12x+2=0#

Divide both sides by #2# to get:

#0 = x^2-6x+1#

#color(white)(0) = x^2-6x+9-8#

#color(white)(0) = (x-3)^2-(2sqrt(2))^2#

#color(white)(0) = ((x-3)-2sqrt(2))((x-3)+2sqrt(2))#

#color(white)(0) = (x-3-2sqrt(2))(x-3+2sqrt(2))#

Hence:

#x = 3+-2sqrt(2)#

Mar 29, 2018

Answer:

#x = 3 +- 2sqrt2#

Explanation:

2x^2 - 12x + 2 = 0
x^2 - 6x + 1 = 0
Use the improved quadratic formula (Socratic search):
#D = d^2 = b^2 - 4ac = 36 - 4 = 32 = 2(16)#
#d = +- 4sqrt2#
There are 2 real roots:
#x = -b/(2a) +- d/(2a) = - (-6/2) +- (4sqrt2)/2#
#x = (3 +- 2sqrt2)#