How do you solve #x^2+6x-5=0# by completing the square?

2 Answers
May 4, 2017

Answer:

#x = 0.742 or x = -6.742#

Explanation:

For a quadratic #ax^2 +bx+c=0#, there are specific conditions required to be able to write it as the square of a binomial.

#a=1, " "(b/2)^2 =c#

#x^2 +6x-5=0#

In this case, #a=1#, but #(6/2)^2 = 9 !=-5#

Move #-5# to the other side and add #9# to both sides.

#x^2 +6x color(blue)(+9) = 5color(blue)(+9)#

#(x+3)^2 = 14" "larr# find the square root of both sides.

#x+3 = +-sqrt14#

#x = +sqrt14-3" or "x = -sqrt14-3#

#x = 0.742 or x = -6.742#

May 4, 2017

Answer:

#x=-3+-sqrt14#

Explanation:

#"to solve using "color(blue)"completing the square"#

add #(1/2"coefficient of the x-term")^2" to both sides"#

#"that is " (6/2)^2=9#

#rArr(x^2+6xcolor(red)(+9))-5=0color(red)(+9)#

#rArr(x+3)^2-5=9#

#"add 5 to both sides"#

#(x+3)^2cancel(-5)cancel(+5)=9+5#

#rArr(x+3)^2=14#

#color(blue)"take the square root of both sides"#

#sqrt((x+3)^2)=color(red)(+-)sqrt14larr" note plus or minus"#

#rArrx+3=+-sqrt14#

#"subtract 3 from both sides"#

#xcancel(+3)cancel(-3)=-3+-sqrt14#

#rArrx=-3+-sqrt14#