How do you solve #\sqrt { 2x + 1} - \sqrt { 4x } = 0#?

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Answer 1 · 2017-05-08T22:10:31

See a solution process below:

First, add #color(red)(sqrt(4x))# to both sides of the equation:

#sqrt(2x + 1) - sqrt(4x) + color(red)(sqrt(4x)) = 0 + color(red)(sqrt(4x))#

#sqrt(2x + 1) - 0 = color(red)(sqrt(4x))#

#sqrt(2x + 1) = sqrt(4x)#

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

#(sqrt(2x + 1))^2 = (sqrt(4x))^2#

#2x + 1 = 4x#

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

#-color(red)(2x) + 2x + 1 = -color(red)(2x) + 4x#

#0 + 1 = (-color(red)(2) + 4)x#

#1 = 2x#

Now, divide each side of the equation by #color(red)(2)# to solve for #x# while keeping the equation balanced:

#1/color(red)(2) = (2x)/color(red)(2)#

#1/2 = (color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2))#

#1/2 = x#

#x = 1/2#

Or

#x = 0.5#