From Förberedande kurs i matematik 1

A first-degree equation which has \displaystyle x=1+\sqrt{3} as a root is \displaystyle x-(1+\sqrt{3}\,)=0, which we can also write as \displaystyle x-1-\sqrt{3} = 0. In the same way, we have that \displaystyle x-(1-\sqrt{3}\,)=0, i.e., \displaystyle x-1+\sqrt{3}=0 is a first-degree equation that has \displaystyle x=1-\sqrt{3} as a root. If we multiply these two first-degree equations together, we get a second-degree equation with \displaystyle x=1+\sqrt{3} and \displaystyle x=1-\sqrt{3} as roots,

\displaystyle (x-1-\sqrt{3}\,)(x-1+\sqrt{3}\,) = 0\,\textrm{.}

The first factor become zero when \displaystyle x=1+\sqrt{3} and the second factor becomes zero when \displaystyle x=1-\sqrt{3}\,.

Nothing really prevents us from answering with \displaystyle (x-1-\sqrt{3}\,)(x-1+\sqrt{3}\,) = 0, but if we want to give the equation in standard form, we need to expand the left-hand side,

\displaystyle \begin{align} (x-1-\sqrt{3}\,)(x-1+\sqrt{3}\,) &= x^{2} - x + \sqrt{3}x - x + 1 - \sqrt{3} - \sqrt{3}x + \sqrt{3} - (\sqrt{3}\,)^{2}\\[5pt] &= x^{2} + (-x+\sqrt{3}x-x-\sqrt{3}x) + (1-\sqrt{3}+\sqrt{3}-3)\\[5pt] &= x^{2}-2x-2 \end{align}

to get the equation \displaystyle x^{2}-2x-2=0\,.

Note: Exactly as in exercise a, we can multiply the equation by a non-zero constant a

\displaystyle ax^{2}-2ax-2a=0

and still have a second-degree equation with the same roots.