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

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Answer 1 · 2016-07-23T11:21:41

$x=4" or "x=36$

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: $36-12-4(color(magenta)(-6))= 48color(red)(" Fails!")$

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