Aus Online Mathematik Brückenkurs 1

We solve the second order equation by combining together the x2- and \displaystyle x -terms by completing the square to obtain a quadratic term, and then solve the resulting equation by taking the root.

By completing the square, the left-hand side becomes

\displaystyle \underline{x^{2}-4x}+3=\underline{\left( x-2 \right)^{2}-2^{2}}+3=\left( x-2 \right)^{2}-1

where the underlined part on the right-hand side is the actual completed square. The equation can therefore be written as

\displaystyle \left( x-2 \right)^{2}-1=0

which we solve by moving the " \displaystyle 1 " on the right-hand side and taking the square root. This gives the solutions

\displaystyle x-2=\sqrt{1}=1 i.e. \displaystyle x=2+1=3

\displaystyle x-2=-\sqrt{1}=-1 i.e. \displaystyle x=2-1=1

Because it is easy to make a mistake, we check the answer by substituting \displaystyle x=1 and \displaystyle x=3 into the original equation.:

\displaystyle x=\text{1}: LHS= \displaystyle 1^{2}-4\centerdot 1+3=1-4+3=0 = RHS

\displaystyle x=3: LHS= \displaystyle 3^{2}-4\centerdot 3+3=9-12+3=0 = RHS