How do you solve #sqrt(4x)=64# and check the solution?

1 Answer
Jul 29, 2017

Answer:

See two solution processes below:

Explanation:

Process 1
First, square both sides of the equation to eliminate the radical while keeping the equation balanced:

#(sqrt(4x))^2 = 64^2#

#4x = 4096#

Now, divide each side of the equation by #color(red)(4)# to solve for #x# while keeping the equation balanced:

#(4x)/color(red)(4) = 4096/color(red)(4)#

#(color(red)(cancel(color(black)(4)))x)/cancel(color(red)(4)) = 1024#

#x = 1024#

Process 2
First, simplify the radical on the left using this rule:

#sqrt(color(red)(a) * color(blue)(b)) = sqrt(color(red)(a)) * sqrt(color(blue)(b))#

#sqrt(4x) = 64#

#sqrt(color(red)(4) * color(blue)(x)) = 64#

#sqrt(color(red)(4))sqrt(color(blue)(x)) = 64#

#2sqrt(color(blue)(x)) = 64#

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

#(2sqrt(color(blue)(x)))/color(red)(2) = 64/color(red)(2)#

#(color(red)(cancel(color(black)(2)))sqrt(color(blue)(x)))/cancel(color(red)(2)) = 32#

#sqrt(color(blue)(x)) = 32#

Now, square both sides of the equation to eliminate the radical and solve for #x# while keeping the equation balanced:

#(sqrt(color(blue)(x)))^2 = 32^2#

#x = 1024#