How do you solve # x^2-14x+5=0# by completing the square?

1 Answer
Mar 21, 2018

Answer:

#x=7-2sqrt11# and #x=7+2sqrt11#

Explanation:

Half the second term of #x# put it in brackets and square it.

#x^2-14x+5-> (x-7)^2+5#

Take away the squared number of the number inside the bracket.

#7^2=49# #therefore# #-49#

#-> (x-7)^2-49+5#

Simplify:

#(x-7)^2-44#

Solving:

Set equal to #0#:

#(x-7)^2-44=0#

Add #44#:

#(x-7)^2=44#

Square root:

#x-7=pmsqrt44#

Add #7#:

#x=7pmsqrt44#

Simplifying:

#sqrt44=sqrt4 xx sqrt11#

#sqrt4=2#

#sqrt11=sqrt11#, It cannot be simplified as it is a prime number.

#therefore# #x=7pmsqrt44 -> x=7pm2sqrt11#

Therefore both answers are:

#x=7-2sqrt11#

and...

#x=7+2sqrt11#