How do you solve #(x-3)^2 = 5# using the quadratic formula?

1 Answer
May 9, 2018

Answer:

#(x-3)^2=5#

#x_1=3+\sqrt(5)#
#x_2=3 -\sqrt(5)#

Explanation:

#(x-3)^2=5#
#x^2-6*x+9=5#
#x^2-6*x+9-5=0#
#x^2-6*x+4=0#

#a*x^2+b*x+c=0#

#x_(1,2)=(-b +- \sqrt(b^2-4*a*c))/(2*a)#

#a=1#
#b=6#
#c=4#

#x_(1,2)=(-(-6) +- \sqrt(6^2-4*1*4))/(2*1)=#
#=(6 +- \sqrt(36-16))/(2*1)=#
#=(6 +- \sqrt(20))/(2)=#
#=(6 +- \sqrt(4*5))/(2)=#
#=(6 +- 2*\sqrt(5))/(2)=#
#=3 +-\sqrt(5)#

#x_1=3+\sqrt(5)#
#x_2=3 -\sqrt(5)#