How do you solve #x^2-6x-25=0# by completing the square?

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Answer 1 · 2017-05-10T16:32:45

#x=3+-sqrt(34)#

First, add #25# to both sides:

#x^2-6x=25#

Now, add something that would make a square on the left side. Use the equation:

#(b/(2a))^2#

#b=-6# and #a=1#. Plugging in gives:

#(-6/2)^2=9#

Add #9# to both sides:

#x^2-6x+9=25+9#

The left side becomes:

#(x-3)^2=34#

Square root both sides:

#x-3=+-sqrt(34)#

Now add 3 to both sides:

#x=3+-sqrt(34)#