Question

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

Answer 1

See the solution process below:

Explanation:

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`