How do you solve #sqrt(5(x+3))=10# and find any extraneous solutions?

1 Answer
Noah G
Jun 2, 2016

Distribute inside the radical, and then square both sides.

Explanation:

#sqrt(5(x + 3)) = 10#

#(sqrt(5x + 15))^2 = 10^2#

#5x + 15 = 100#

#5x = 85#

#x = 17#

Since this is a radical equation, we must always check our answer back in the original equation to verify that there are no extraneous solutions. Checking, we find that #x = 17# satisfies our equation; the solution set is #{x = 17}#.

Hopefully this helps!

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