How do you solve #4(x-5)^(1/2)=28#?

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Answer 1 · 2017-05-27T12:07:13

See the solution process below:

First, divide each side of the equation by #color(red)(4)# to isolate the term with the exponent while keeping the equation balanced:

#(4(x - 5)^(1/2))/color(red)(4) =28/color(red)(4)#

#(color(red)(cancel(color(black)(4)))(x - 5)^(1/2))/cancel(color(red)(4)) =7#

#(x - 5)^(1/2) =7#

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

#((x - 5)^color(red)(1/2))^color(blue)(2) = 7^2#

#(x - 5)^(color(red)(1/2)xxcolor(blue)(2)) = 49#

#(x - 5)^1 = 49#

#x - 5 = 49#

Now, add #color(red)(5)# to each side of the equation to solve for #x# while keeping the equation balanced:

#x - 5 + color(red)(5) = 49 + color(red)(5)#

#x - 0 = 54#

#x = 54#