Question

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

Answer 1

`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!")`