"Question: a) express x^2 - 4x + 1 in the form (x + a )^2 + b"

This is called "completing the square". You should have learned, before you could be expected to do a problem like this, that \(\displaystyle (x+ a)^2= x^2+ 2ax+ a^2\).

(You can get that by actually doing the multiplication: \(\displaystyle (x+ a)^2= (x+ a)(x+ a)= x(x+ a)+ a(x+ a)= x^2+ ax+ ax+ a^2= x^2+ 2ax+ a^2\).)

Compare the first part of that, "\(\displaystyle x^2+ 2ax\)", to your "\(\displaystyle x^2- 4x\)". They will be exactly the same if \(\displaystyle 2a= -4\) which means that \(\displaystyle a= -2\). Yes, \(\displaystyle (x- 2)^2= x^2- 4x+ 4\). In order to "complete the square", to make \(\displaystyle x^2-4x\) a complete square, you need to add 4. Of course you **can't** just "add 4", that would change the value. But you **can** do the old trick of "add" and "subtract" the same thing so that you change the "form" but not the value: \(\displaystyle x^2- 4x+ 1= x^2- 4x+ 4- 4+ 1= (x^2- 4x+ 4)+ (-4+ 1)= (x- 2)^2- 3\). That is "in the form (x+ a)^2+ b" with a= -2 and b= -3.

For (b), write \(\displaystyle x^2- 4x+ 1= (x- 2)^2- 3= 0\). Adding 3 to both sides, \(\displaystyle (x- 2)^2= 3\). Taking the square root of both sides, \(\displaystyle x- 2= \pm\sqrt{3}\) so, adding 2 to both sides, \(\displaystyle x= 2\pm\sqrt{3}\).