
theorem
  for a,b,c,x,y being Real st a <> 0 & b^2-4*a*c + 8*a^2 > 0 & y
= x + 1/x holds Polynom(a,b,c,b,a,x) = 0 implies for y1,y2 being Real st
  y1 = (-b+sqrt(b^2-4*a*c+8*a^2))/(2*a) & y2 = (-b-sqrt(b^2-4*a*c+8*a^2))/(2*a)
holds x <> 0 & (x = (y1 + sqrt delta(1,(-y1),1))/2 or x = (y2 + sqrt delta(1,(-
y2),1))/2 or x = (y1 - sqrt delta(1,(-y1),1))/2 or x = (y2 - sqrt delta(1,(-y2)
  ,1))/2)
proof
  let a,b,c,x,y be Real;
  assume that
A1: a <> 0 and
A2: b^2-4*a*c+8*a^2 > 0 and
A3: y = x + 1/x and
A4: Polynom(a,b,c,b,a,x) = 0;
A5: x <> 0 by A1,A3,A4,Th2;
  set f = c - 2*a;
  a*y^2+ b*y + c - 2*a = 0 by A1,A3,A4,Th2;
  then a*y^2+ b*y + (c - 2*a) = 0;
  then
A6: Polynom(a,b,f,y) = 0 by POLYEQ_1:def 2;
  let y1,y2 be Real;
  assume
A7: y1 = (-b+sqrt(b^2-4*a*c+8*a^2))/(2*a) & y2 = (-b-sqrt(b^2-4*a*c+8*a
  ^2))/(2*a );
  x*y = x^2 +(x*(1/x)) by A3;
  then x*y + 0 = (x^2 + 1) by A5,XCMPLX_1:106;
  then 1*x^2+((-y)*x) + 1 = 0;
  then
A8: Polynom (1,-y,1,x) = 0 by POLYEQ_1:def 2;
  delta(1,(-y),1) = (-y)^2 -4*1*1 by QUIN_1:def 1
    .= (x^2 + (2*(x*(1/x))) +(1/x)^2)-4 by A3
    .= (x^2 +2*1 +(1/x)^2) -4 by A5,XCMPLX_1:106
    .= x^2 + (-2*1 +(1/x)^2)
    .= x^2 + (-2*(x*(1/x)) +(1/x)^2) by A5,XCMPLX_1:106
    .= ( x - (1/x) )^2;
  then
A9: x = (-(-y) + sqrt delta(1,(-y),1))/(2*1) or x = (-(-y) - sqrt delta(1,(
  -y),1))/(2*1) by A8,POLYEQ_1:5,XREAL_1:63;
A10: b^2 - 4*a*f = b^2 - 4*a*c + 8*a^2;
  then delta(a,b,f) > 0 by A2,QUIN_1:def 1;
  then y = (-b + sqrt delta(a,b,f))/(2*a) or y = (-b - sqrt delta(a,b,f))/(2*
  a) by A1,A6,POLYEQ_1:5;
  then y = y1 or y = y2 by A7,A10,QUIN_1:def 1;
  hence thesis by A1,A3,A4,A9,Th2;
end;
