How do you solve #x^2-2x-1 = 0# ?

1 Answer
George C.
Jun 18, 2017

#x=1+sqrt(2)" "# or #" "x=1-sqrt(2)#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

We can use this with #a=(x-1)# and #b=sqrt(2)# as follows:

#0 = x^2-2x-1#

#color(white)(0) = x^2-2x+1-2#

#color(white)(0) = (x-1)^2-(sqrt(2))^2#

#color(white)(0) = ((x-1)-sqrt(2))((x-1)+sqrt(2))#

#color(white)(0) = (x-(1+sqrt(2)))(x-(1-sqrt(2)))#

Hence:

#x=1+sqrt(2)" "# or #" "x=1-sqrt(2)#

#color(white)()#
Notes

So what did we do?

Given:

#x^2-2x-1#

we separated this expression into a difference of squares.

For the left hand square we used:

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

Note that in general:

#(x+c)^2 = x^2+2cx+c^2#

In our example #2c = -2# to match the coefficient of #x#, so we needed #c=-1#

Then we had:

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

To be viewed as a difference of squares, we need that final #2# to be the square of something. That something is #sqrt(2)#. (We could have picked #-sqrt(2)# instead, but it makes no difference).

Having got our difference of squares, we factored into linear factors and hence found our zeros.

Related questions
Impact of this question
735 views around the world
You can reuse this answer
Creative Commons License