مسلمانوں کا نیوٹن ۔ ڈاکٹر عبدالسلام

مسلمانوں کا نیوٹن ۔ ڈاکٹر عبدالسلام — Page 371

361 ABDUS SALAM 5۔We know S₂+p S₂+p S₁ + 3p₁ = 0 Substituting for S۔S₁, S₁۔P₂ we get 6p₁ = 3p₁² + p₁³ - 10 ap₁ +2 Substitute for p、 from (g) in (ƒ) we get P₁' - (4a - 3) pr² - 4p₁ = 0 P₁ {p₁³ - (4a − 3) ₪ − 4} = 0 - : (g) (k) The cubic in p₁ can be solved by the usual methods۔pɩ known, På۔Pr۔Pɩ can easily be discovered and the corresponding biquadratic in can be framed۔The hiquadratic may be solved by usual methods for x, y, zu; For the particular value p₁=0 the biquadratic is - t* − 2a t³ + } l'+ a³ − a=0 6۔By employing the same methods, we can solve the system of equations ³=a+y y=a+'z 2'-a+r much more rapidly than Ramanujan did۔His is a very laborious method۔Govt۔College, Lahore۔} ABDUS SALAM, Fourth Year Student