# Format of the scenario data file: Multi-record data file.
#
# Record 1: header.
#	Line 1: title
#	Line 2: author
#	Line 3: email
#	Line 4: translator
#	Line 5: email
#	Line 6: format (html,tex; default html)
#	Line 7 and up: random data.
# Record 2: presentation of the problem.
# Record 3: Good scenario. One step per line.
# Record 4: Seemingly bad reason(s) for each step, one line per step.
# Record 5: Remarks. One line per step.
# Record 6: Reserved.
# Record 7 and up: Bad scenarios.
#	Line 1: starting step, bad reason.
#	Line 2: remark.
#	Line 3 and up: one step per line.
#

:Quadratic I
XIAO, Gang
xiao@unice.fr


html
&gt;,$m_ge@&lt;,$m_le

:Here is an argument to solve the inequality
 x<sup>2</sup> $r1 -2x.
:Moving the term -2x to the left, the inequality becomes x<sup>2</sup>+2x $r1 0.
 Adding 1 bo both sides, x<sup>2</sup>+2x+1 $r1 1.
 The left side is now a square: (x+1)<sup>2</sup> $r1 1.
 It is therefore equivalent to: -1 $r1 x+1 or x+1 $r1 1.
 Add -1 to each side: -2 $r1 x or x $r1 0.
:add_neg2, add_sign
 add_sign
 alg_err
 square
 illegal, add_sign, add_neg2
:  




:

: 1, div_neg
 What happens if x is negative?
 Dividing both sides by x, x $r1 -2.
: 1, add_sign

 Moving the term 2x to the left, the inequality becomes x<sup>2</sup>-2x $r1 0.
 Adding 1 bo both sides, x<sup>2</sup>-2x+1 $r1 1.
 The left side is now a square: (x-1)<sup>2</sup> $r1 1.
 It is therefore equivalent to: -1 $r1 x-1 or x-1 $r1 1.
 Add 1 to each side: 0 $r1 x or x $r1 2.
: 1, add_sign, 4, square

:Moving the term 2x to the left, the inequality becomes x<sup>2</sup>-2x $r1 0.
 Adding 1 bo both sides, x<sup>2</sup>-2x+1 $r1 1.
 The left side is now a square: (x-1)<sup>2</sup> $r1 1.
 It is therefore equivalent to x-1 $r1 1.
 Moving -1 to the right, x $r1 2.
: 2, add_sign
 One must add the SAME term to both sides of the inequality.
 Adding 1 then subtracting it back, x<sup>2</sup>+2x+1 $r1 -1.
 The left side is now a square: (x+1)<sup>2</sup> $r1 -1.
 The square of a real number being always $r1 -1, any value of x is a solution.
: 3, alg_err

 The left side is now a square: (x+2)<sup>2</sup> $r1 1.
 It is therefore equivalent to: -1 $r1 x+2 or x+2 $r1 1.
 Add -2 to each side: -3 $r1 x or x $r1 -1.
:4, square
 Think of the case where x+1 is negative!
 It is therefore equivalent to x+1 $r1 1.
 Moving +1 to the right, x $r1 0.
:4, square, 5, add_sign
 Think of the case where x+1 is negative!
 It is therefore equivalent to x+1 $r1 1.
 Moving +1 to the right, x $r1 2.
:4, bad_chain
 The chain of inequalities imply -1 $r1 1, which is never true.
 It is therefore equivalent to: -1 $r1 x+1 $r1 1.
 Add -1 to each side: -2 $r1 x $r1 0.
:4, bad_chain, 5, add_neg
 The chain of inequalities imply -1 $r1 1, which is never true.
 It is therefore equivalent to: -1 $r1 x+1 $r1 1.
 Add -1 to each side: -2 $r2 x $r2 0.
