If #\sqrt { n ^ { 2} + 31} - 2y = 0 and y = 8#, what is the value of #n#?

1 Answer
Nov 22, 2017

See a solution process below:

Explanation:

Substitute #color(red)(8)# for #color(red)(y)# in the equation and solve for #n#:

#sqrt(n^2 + 31) - 2color(red)(y) = 0# becomes:

#sqrt(n^2 + 31) - (2 xx color(red)(8)) = 0#

#sqrt(n^2 + 31) - 16 = 0#

First, add #color(red)(16)# to each side of the equation to isolate the radical while keeping the equation balanced:

#sqrt(n^2 + 31) - 16 + color(red)(16) = 0 + color(red)(16)#

#sqrt(n^2 + 31) - 0 = 16#

#sqrt(n^2 + 31) = 16#

Next, square both sides of the equation to eliminate the radical while keeping the equation balanced:

#(sqrt(n^2 + 31))^2 = 16^2#

#n^2 + 31 = 256#

Then, subtract #color(red)(31)# from each side of the equation to isolate the #n# term while keeping the equation balanced:

#n^2 + 31 - color(red)(31) = 256 - color(red)(31)#

#n^2 + 0 = 225#

#n^2 = 225#

Now, take the square root of each side of the equation to solve for #n# while keeping the equation balanced. Remember, the square root of a number produces both a positive and negative result:

#sqrt(n^2) = +-sqrt(225)#

#n = +-15#

#n = {-15, 15}#