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

1 Answer
Jul 26, 2018

Answer:

#x=1 or x=-6#

Explanation:

Here,
#x^2+5x-6=0#
...........................................................................................................................
To complete perfect square let #M# be the third term,
such that #x^2+5x +M # is a perfect square.

So,

#color(blue)((i)1^(st) term=x^2#
#color(blue)((ii)2^(nd)term=5x#
#color(blue)((iii)3^(rd)term=M#

Formula :
#color(blue)(3^(rd)term=(2^(nd)term)^2/(4xx 1^(st) term)#

#M=(5x)^2/(4xxx^2)=(25x^2)/(4x^2)=25/4#

#i.e. x^2+5x+(25/4) # is a perfect square.
......................................................................................................................
So,

#x^2+5x=6#

#=>x^2+5x+color(red)(25/4)=6+color(red)(25/4#

#=>(x+5/2)^2=(49)/4=(7/2)^2#

#=>x+5/2=+-7/2#

#=>x+5/2=7/2 or x+5/2=-7/2#

#=>x=7/2-5/2 or x=-7/2-5/2#

#=>x=2/2 or x=-12/2#

#=>x=1 or x=-6#