How do you solve and check for extraneous solutions in $3sqrt(2x+4)=12$ ?

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Answer 1 · 2015-08-10T10:23:57

$x = 6$

First, isolate the radical on one side of the equation by dividing both sides by

$(color(red)(cancel(color(black)(3))) * sqrt(2x+4))/color(red)(cancel(color(black)(3))) = 12/3$

$sqrt(2x+4) = 4$

Square both sides of the equation to get rid of the square root

$(sqrt(2x+4))^2= 4^2$

$2x+4 = 16$

Finally, isolate $x$ on one side of the equation by adding $-4$ to both sides and dividing everything by $2$

$2x + color(red)(cancel(color(black)(4))) - color(red)(cancel(color(black)(4))) = 16 - 4$

$(color(red)(cancel(color(black)(2)))x)/color(red)(cancel(color(black)(2))) = 12/2$

$x = color(green)(6)$

How do you solve and check for extraneous solutions in 3sqrt(2x+4)=123sqrt(2x+4)=12?