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How do you solve and check for extraneous solutions in #x^(1/2)+3=4#?

1 Answer

Aug 5, 2015

I found #x=1#

Explanation:

You have:
#sqrt(x)+3=4#
#sqrt(x)=4-3=1# square both sides:
#x=1#
substituting back into the original equation:
#x=1# then: #sqrt1+3=4# ok!

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