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

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Answer 1 · 2017-02-27T19:02:07

See the entire solution process below:

First, square each side of the equation:

#(sqrt(x))^2 = (2x)^2#

#x = 4x^2#

Next, subtract #color(red)(x)# from each side of the equation:

#x - color(red)(x) = 4x^2 - color(red)(x)#

#0 = 4x^2 - x#

#4x^2 - x = 0#

Then, factor an #x# out of each term on the left side of the equation:

#x(4x - 1) = 0#

Now, solve each term for #0# to find all the solutions to the problem:

Solution 1)

#x = 0#

Solution 2)

#4x - 1 = 0#

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

#4x - 0 = 1#

#4x = 1#

#(4x)/color(red)(4) = 1/color(red)(4)#

#(color(red)(cancel(color(black)(4)))x)/cancel(color(red)(4)) = 1/4#

#x = 1/4#

The solution is: #x = 0# and #x = 1/4#