Question

How do you solve `x=5+(3x-11)^(1/2)`?

Answer 1

`x=9" or " x=4`

Explanation:

recall: `x^(1/2) = sqrtx`

`color(white)(xxxxxx.xxxxxxxx)x-5 = (3x-11)^(1/2)`

`color(white)(xxxxxxxxxxxxx)(x-5)^2 = sqrt(3x-11)^2" " larr` square

`color(white)(xxxxxxxxx)x^2-10x +25 = 3x -11`

`color(white)(xx)x^2 -10x -3x +25 +11 = 0`

`color(white)(xxxxxxxxx)x^2 -13x +36 =0" "larr` factor

`color(white)(xxxxxxxxx)(x-9)(x-4) = 0`

If `x-9 = 0 " "rarr x = 9`
If `x-4 =0 " "rarr x= 4`