How do you solve #sqrt(x+72) = x#?

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Answer 1 · 2018-03-24T08:59:57

#x=9#

Square both sides and place terms on one side to get a more familiar quadratic equation:

#x+72 = x^2#

#0=x^2 -x - 72#

It factors to

#(x+8)(x-9)=0#

So the roots are #-8# and #9#.

However, we must plug the answers back into the original equation and check.

We see that #-8# will not work, because only the principal square root is used.

#sqrt(-8+72) = sqrt 64 = 8#.

#8 != -8" "larr 8# is an extraneous solution.

So #x=9# is the only solution:

#sqrt(9+72) = sqrt81 = 9#

#9=9#