reasoning in quadratic equations with one unknown - Cerme 7

“Find the solution set of the equation “ 2 − = 12”Figure 3: An example of students’ mistake when just the form of equation changed.In the first type of wrong solution (see Figure 1), students carried the term -2x fromleft side to the right, and then simplified the term x in both sides of the equation.Consequently, they ignored one of the roots of the equation, which is 0. In the secondtype of wrong solution (see Figure 2), students tried to factorize the equation. Here,students perceived the form ax 2 -bx=0 just like ax 2 +bx+c=0 and thought -2x as theconstant term of the quadratic equations. Even, when just the form of the equationwas changed instead of the structure (e.g., ax 2 +bx=c where a, b, c≠0), 12% of thestudents incorrectly solved the quadratic. Because the constant term was in the rightside, they didn’t perceive that the equation was in standard form (see Figure 3). Inthis type of solution, they were able to find only one of the roots, 4.StatementsQuestion 5To solve the equation“(x-3)∙(x-2) = 0”for real numbers, Alianswered in a singleline that:“x=3 or x=2”Is this answercorrect? If it is correct,how can you show itcorrectness?Students’ types of responses with their reasoningI. II. III. IV.“The answer isRight”Since I wrote(x-3)∙(x-2) = 0as x 2 -5x+6=0and then I factorizeto find roots of it.from (x-3)=0 andfrom (x-2)=0“x=3 and x=2”Table 1: Common examples of students’ types of responses with their reasoning forquestion 5.Although all of the students stated that Ali’s answer was correct by choosing eitherone of the statements I, II, III, and IV, the ways they justified for the correctness ofAli’s solution were different. For instance, in statement I, students first transformedthe factorized expression into the standard form, and then factorized the expressionagain in the same way and found the roots by rote. In statement II, studentsunconsciously applied the null factor law. In statement III, the way of justification forsolution was based on substitution method. In all of these three statements, they couldnot clearly justify the correctness of the solution. In statement IV, students substitutedx=3 into (x-3) and x=2 into (x-2) simultaneously, and concluded that their solution5“The answeris Right”Because(x-3)=0(x-2)=0x=3x=2“Theanswer isRight”.Since wesubstitute“3 and 2”into x, theequation isprovided.(explanationmade onlywith words)“The answer isRight”.If the x=2and x=3 aresubstituted intothe equation(3-3)⋅ (2-2)=00.0 = 0