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

1 Answer
EZ as pi
Sep 19, 2016

#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#

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