How do you solve #x^2-6x+11=0# by completing the square?

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Answer 1 · 2018-06-25T22:09:12

#x=3+-isqrt2#

Let's first subtract #11# from both sides to get

#x^2-6x=-11#

When we complete the square of an equation of the form

#ax^2+bx=-c#

We take half of our #b# value, square it, and add it to both sides.

Our #b# value is #-6#, half of that is #-3#, and that value squared is #9#. Adding it to the left and right gives us

#color(steelblue)(x^2-6x+9)=-11+9#

What I have in blue can be factored as #(x-3)^2#. This gives us

#(x-3)^2=-2#

Taking the square root of both sides gives us

#x-3=sqrt(-2)#

Which can be rewritten as

#x-3=sqrt(-1)*sqrt2#

Note that #i=sqrt(-1)#. With this definition in mind, we now have

#x-3=+-isqrt2#

Adding #3# to both sides gives us

#x=3+-isqrt2#

Hope this helps!