How do you solve #x^2+30x-7=0# by completing the square?

1 Answer
Aug 19, 2016

#x = 0.2315 or x = -30.2315#

Explanation:

The method using completing the square is based on:

#(x-y)^2 = x^2 - 2xy +y^2#

#(x-6)^2 = x^2 color(red)(-12)x +color(blue)(36) # Note that: #(color(red)(-12)/2)^2 = color(blue)(36)#
This relationship always exists in squaring a binomial.

#x^2 +30x-7 =0" 7 is not the correct constant"#

Move the 7 to the other side and add in the correct constant on both sides.
#x^2 color(red)(+30)x color(blue)(+225)= 7 color(blue)(+225)" "(color(red)(30)/2)^2 = color(blue)225#
#(x+15)^2 = 232" "x^2 +30x+225" is a square"#

#x + 15 = +-sqrt232 " square root both sides"#

Solve for x twice:

#x = +sqrt232 -15 = 0.2315#

#x = - sqrt232 -15 =-30.2315#