Aus Online Mathematik Brückenkurs 1

A first-degree equation which \displaystyle x=\text{ 1}+\sqrt{\text{3}} as a root is \displaystyle x-\left( \text{1}+\sqrt{\text{3}} \right)=0, which we can also write as

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

\displaystyle \left( x-\text{1-}\sqrt{\text{3}} \right)\left( x-\text{1+}\sqrt{\text{3}} \right)=0

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

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

\displaystyle \begin{align} & \left( x-\text{1-}\sqrt{\text{3}} \right)\left( x-\text{1+}\sqrt{\text{3}} \right)=x^{2}-x+\sqrt{3}-x+1-\sqrt{3}-\sqrt{3}x+\sqrt{3}-\left( \sqrt{3} \right)^{2} \\ & =x^{2}+\left( -x+\sqrt{3}x-x-\sqrt{3}x \right)+\left( 1-\sqrt{3}+\sqrt{3}-3 \right) \\ & =x^{2}-2x-2 \\ \end{align}

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

NOTE: Exactly as in exercise \displaystyle \text{a}, we can multiply the equation by a non-zero constant \displaystyle a,

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

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