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Aus Online Mathematik Brückenkurs 1

The equation can be written in normalized form (i.e. the coefficient in front of x2 is 1 ) by dividing both sides by 4,

x2−7x+413=0

Completing the square on the left-hand side,

x2−7x+413=x−272−272+413=x−272−449+413=x−272−436=x−272−9

The equation can therefore be written as

x−272−9=0 

and taking the square root gives the solutions as

x−27=9=3  i.e. x=27+3=213

x−27=−9=−3  i.e. x=27−3=21

As an extra check, we substitute x=1/2 and x=13/2 into the equation:

x=12: LHS =4212−2821+13=441−14+13=1−14+13=  RHS

x=132: LHS =42132−28213+13=44169−1413+13=169−182+13=  RHS