How do you solve #x^2+10x-1=0# by completing the square?

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Answer 1 · 2018-04-07T18:32:46

#x=+-sqrt26-5#

Isolate all terms involving #x# on one side and move the constant to the other side.

#x^2+10x=1#

Now, the original quadratic, #x^2+10x-1,# is in the form #ax^2+bx+c# where #a=1, b=10, c=-1#.

To complete the square, we first want to add #(b/2)^2# to each side.

#(b/2)^2=(10/2)^2=5^2=25#, so we add #25# to each side.

#x^2+10x+25=1+25#

The left side needs to be factored. Fortunately, since we added #(b/2)^2# to each side, the factored form will simply be #(x+b/2)^2=(x+5)^2#

#(x+5)^2=26#

Now, take the root of both sides, accounting for positive and negative answers on the right.

#sqrt((x+5)^2)=+-sqrt26#

#x+5=+-sqrt26#

Solving for #x# yields

#x=+-sqrt26-5#