Processing Math: Done

From Förberedande kurs i matematik 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−22−22+3=x−22−1 

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

x−22−1=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