How do you solve #sqrtx+12=x#?

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How do you solve #sqrtx+12=x#?

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Answer 1 · 2015-04-18T18:55:44

Take #12# to the right and then square both sides to get:
#x=(x-12)^2#
#x=x^2-24x+144#
#x^2-25x+144=0#
Solve for #x# and then check the results (you´ll get 2 of which only one acceptable) by substituting back into your original equation.

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How do you solve #sqrtx+12=x#?

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