How do you solve the equation #x^2+4x+4=25# by completing the square?

1 Answer
Dec 29, 2016

Answer:

#3# or #-7#. Re-write the LHS as #(x+2)^2# then take the square root of both sides. That tells you that #x=+5-2# or #-5-2#.

Explanation:

The way the question is set out half-does the problem for you, as "the square is completed" already. If the question had been "solve #x^2+4x-21=0#, you would have taken the #21# to the RHS, then added the square of half the number in front of the #x# to both sides (called "completing the square').. In this case, the number in front of the #x# is #4#, half it is #2#, square it is #2^2=4#, so you get the equation that you stated.

This may be re-written as #(x+2)^2=25#.

So #x+2 = +-5#.

So #x 3# or #-7#.