How do you solve #-3=(37-3n)^(1/2)-n#?

1 Answer
Nityananda
Sep 19, 2016

n = 7, - 4

Explanation:

-3 = #(37 - 3n)^(1/2) -n #
or, -3 + n = #(37 - 3n)^(1/2)# [squaring both sides]
or, #(-3 + n)^2 = (37 - 3n)^[(1/2)*2]#
or, 9 - 6n + #n^2# = 37 - 3n
or, #n^2 - 3n - 28 = 0 #
or, #n^2 -7n + 4n - 28 = 0#
or, n ( n - 7 ) + 4 ( n - 7 ) = 0
or, (n - 7) (n + 4 ) = 0
or, n = 7, - 4

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