How do you solve #sqrt(8x) = sqrt(4+2x)#?

Question

1124 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-02T12:47:11

See a solution process below:

First, square both sides of the equation to eliminate the radicals while keeping the equation balanced:

#(sqrt(8x))^color(red)(2) = (sqrt(4 + 2x))^color(red)(2)#

#8x = 4 + 2x#

Next, subtract #color(red)(2x)# from each side of the equation to isolate the #x# term while keeping the equation balanced:

#8x - color(red)(2x) = 4 + 2x - color(red)(2x)#

#(8 - color(red)(2))x = 4 + 0#

#6x = 4#

Now, divide each side of the equation by #color(red)(6)# to solve for #x# while keeping the equation balanced:

#(6x)/color(red)(6) = 4/color(red)(6)#

#(color(red)(cancel(color(black)(6)))x)/cancel(color(red)(6)) = 2/3#

#x = 2/3#