How do you solve #sqrt(x+12)=x#?

1 Answer
May 19, 2015

#sqrt(x+12) =x#

squaring both the sides
#(sqrt(x+12))^2 =x^2#
#x+12 = x^2#
# x^2 -x -12 =0#

we can find the roots by Splitting the Middle Term of this expression to factorise it:

# x^2 -x -12 = x^2 -4x +3x -12 #
#= x(x-4)+3 (x-4)#
#= (x+3)(x-4)#

the solution is #color(blue)x=-3 , color(blue)x=4#