How do you solve #5^{2}=\sqrt{2x}^{2}#?

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Answer 1 · 2017-12-06T04:46:22

See solution process below:

First, square each side of the equation giving (Remember, a square root squared is equal to the term within the square root or #sqrt(a)^2 = a#):

#25 = 2x#

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

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

#25/2 = (color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2))#

#25/2 = x#

#x = 25/2# or #x = 12.5#