Processing Math: Done

Aus Online Mathematik Brückenkurs 1

We solve the second order equation by combining together the x²- and 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

x2−4x+3=(x−2)2−22+3=(x−2)2−1,

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

(x−2)2−1=0

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

• x−2=1=1   i.e. x=2+1=3

• x−2=−1=−1   i.e. x=2−1=1.

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

• x = 1:  LHS=12−41+3=1−4+3=0=RHS,

• x = 3:  LHS=32−43+3=9−12+3=0=RHS.