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:

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

$x - 0 = 54$

$x = 54$

How do you solve 4(x-5)^(1/2)=284(x-5)^(1/2)=28?