Lösung 2.3:4b

Aus Online Mathematik Brückenkurs 1

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

(x−1−3)(x−1+3)=0.

The first factor become zero when x=1+3  and the second factor becomes zero when x=1−3 .

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

(x−1−3)(x−1+3)=x2−x+3x−x+1−3−3x+3−(3)2=x2+(−x+3x−x−3x)+(1−3+3−3)=x2−2x−2

to get the equation x2−2x−2=0.

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

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