How do you solve #x-12=sqrt(16x)#?

1 Answer
Jul 23, 2016

Answer:

#x=4" or "x=36#

Explanation:

What you do to 1 side of the equation you do to the other side.

Lets 'get rid' of the square root.

Square both sides:

#(x-12)^2=16x#

multiplied out the brackets
#x^2-24x+144" "=" "16x#

Subtract #color(blue)(16x)# from both sides
#color(brown)(x^2-24xcolor(blue)(-16x)+144" "=" "16xcolor(blue)(-16x))#

#x^2-40x+144" "=" "0#

Note that #(-4)xx(-36)=+144"; "-4-36=40#

#(x-4)(x-36)" "=" "0#

#x=4" or "x=36#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check #x-12-4sqrt(x)=0#

,................................................
At #x=4 ->(4)-12-4(color(magenta)(+-2)) #

Try: #(4)-12-4(color(magenta)(+2)) = -16 color(red)(" Fail!")#
Try: #(4)-12-4(color(magenta)(-2)) = 0 color(green)(" Works!")#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
At #x=36 -> 36-12-4(color(magenta)(+-6))#

Try: #36-12-4(color(magenta)(+6))= 0color(green)(" Works!")#
Try: #36-12-4(color(magenta)(-6))= 48color(red)(" Fails!")#