# 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
&lt;, $m_le@&gt;, $m_ge
: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 $r1 1.
 Add 1 to each side: 0 $r1 x $r1 2.
:add_neg2, add_sign
 add_sign
 alg_err
 square, bad_chain
 illegal, add_sign, add_neg2, bad_chain
:  


(x-1)<sup>2</sup> $r1 1 is indeed equivalent to {-1 $r1 x-1 and x-1 $r1 1}, which can be written as -1 $r1 x-1 $r1 1.
We have indeed the right to add a same number to each side of a chain of inequalities.
:

: 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 $r1 1.
 Add -1 to each side: -2 $r1 x $r1 0.
: 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.
 Add -1 to each side: x $r1 0.
: 1, add_sign, 4, square, 5, add_neg

 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.
 Add -1 to each side: 0 $r1 x $r2 0.
: 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.
 As the square can never be negative, there is no 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 $r1 1.
 Add 2 to each side: 1 $r1 x $r1 3.
: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 2.
: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 0.
:4, square

 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.
:4, square, 5, add_sign

 It is therefore equivalent to: -1 $r1 x-1 or x-1 $r1 1.
 Move the term -1 to other sides: -2 $r1 x or x $r1 0.
:5, illegal
 One must add the same term to each side of the chain.
 Add 1 to the first inequality: 0 $r1 x $r1 1.
:5, illegal
 One must add the same term to each side of the chain.
 Move -1 to the left: -2 $r1 x $r1 1.
:5, illegal
 One must add the same term to each side of the chain.
 Move -1 to the right: -1 $r1 x $r1 0.
:5, add_sign
 One must add the same term to each side of the chain.
 Move -1 to other sides: -2 $r1 x $r1 0.

