How do you solve #x^2-2x-14=0# by completing the square?

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Answer 1 · 2017-07-27T02:54:30

#x= +-sqrt(15)+1#

First, you have to move the 14 to the other side by adding it :

#x^2-2x=14#

To complete the square you divide the term with one x by 2 and then take that number and square it :

#2/2=1# , #1^2=1#

We then take that 1 and add it to both sides of the equation :

#x^2-2x+1=15#

Now we can factor the left side of the equation :

#(x-1)^2=15#

Take the square root of both sides to get rid of the square on the left hand side (be sure to add the #+-# on the 15) :

#sqrt((x-1)^2)=+-sqrt(15)#

which then equals...

#x-1=+-sqrt(15)#

Now get x by itself :

#x= +-sqrt(15)+1#