Solver COMPLETING THE SQUARE solver for quadratics

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==section input

Solve for variable *[input X=x]: *[input a=1]x2+*[input b=-1]x+*[input c=-2] = 0;

==section solution
perl

print "Read <A HREF=Fool-Proof-Completing-the-Square-Part-I.lesson>this lesson on completing the square by prince_abubu</A>, if you do not know how to complete the square. ";

print "Let's convert {{{$a$X^2+$b$X+$c=0}}} to standard form by dividing both sides by $a: \n\n";
$c = $c/$a;
$b = $b/$a;
$a = 1;

print "We have: {{{$a$X^2+$b$X+$c=0}}}. \nWhat we want to do now is to change this equation to a complete square {{{($X+somenumber)^2 + othernumber}}}. How can we find out values of somenumber and othernumber that would make it work?\n\n";

$somenumber = $b/2;

print "Look at {{{($X+somenumber)^2}}}: {{{($X+somenumber)^2 = $X^2+2*somenumber*x + somenumber^2}}}. Since the coefficient in our equation {{{$a$X^2+highlight_red( $b) * $X+$c=0}}} that goes in front of $X is $b, we know that $b=2*somenumber, or {{{somenumber = $b/2}}}. So, we know that our equation can be rewritten as {{{($X+$b/2)^2 + othernumber}}}, and we do not yet know the other number.\n\n";

my $othernumber = $c-$b*$b/4;

print "We are almost there. Finding the other number is simply a matter of not making too many mistakes. We need to find 'other number' such that  {{{($X+$b/2)^2 + othernumber}}} is equivalent to our original equation {{{$a$X^2+$b$X+highlight_green( $c )=0}}}.

{{{($X+$b/2)^2 + othernumber = $X^2+$b$X+highlight_red( $b^2/4+othernumber )}}}
The highlighted red part must be equal to $c (highlighted green part).

{{{$b^2/4 + othernumber = $c}}}, or {{{othernumber = $c-$b^2/4 = $othernumber}}}.\n\n";

my $mothernumber = -$othernumber;

print "So, the equation converts to {{{($X+$b/2)^2 + $othernumber = 0}}}, or {{{($X+$b/2)^2 = $mothernumber}}}.

Our equation converted to a square {{{($X+$b/2)^2}}}, equated to a number ($mothernumber).

if( $mothernumber > 0 ) {
  print "Since the right part $mothernumber is greater than zero, there are two solutions:

{{{system( ($X+$b/2) = +sqrt( $mothernumber ), ($X+$b/2) = -sqrt( $mothernumber ) )}}}
  ";
  my $s = sqrt( $mothernumber );
print ", or

{{{system( ($X+$b/2) = $s, ($X+$b/2) = -$s )}}}
   {{{system( $X+$b/2 = $s, $X+$b/2 = -$s )}}}
   {{{system( $X = $s-$b/2, $X = -$s-$b/2 )}}}
  
";

($x1, $x2) = ($s-$b/2, -$s-$b/2);

print "   {{{system( $X = $x1, $X = $x2 )}}}\n\n";
  print "Answer: $X=$x1, $x2.\n";

} elsif( $mothernumber == 0 ) {
  ($x1, $x2) = (-$b/2, -$b/2);
  print "Since the right part is zero, we simply have {{{$X-$b/2 = 0}}}, or {{{$X = $b/2}}}, or $X=$x1";
} else {
  print "There is no number whose square can be negative. So, there is no solution to this equation";
}

==section output
x1
x2
==section check

X=x a=1 b=0 c=0 x1=0 x2=0

==section practice
perl

$X=(qw(x y z q))[randint(0,3) ];
  my $x1 = randint( -4, 4 );
  my $x2 = randint( -4, 4 );
  $a=1;
  $b = -($x1+$x2);
  $c = $x1*$x2;