Để phương trình có 2 nghiệm phân biệt thì \(\Delta > 0\) ⇔ a2 – 4b > 0
Theo hệ thức Viet, ta có:
\(\left\{ {\begin{array}{*{20}{c}}{{x_1} + {x_2} = - a}\\{{x_1}.{x_2} = b}\end{array}} \right.\)
Ta có: \({x_1}^3 - {x_2}^3 = 35\)
⇔ \(\left( {{x_1} - {x_2}} \right)\left( {{x_1}^2 + {x_1}{x_2} + {x_2}^2} \right) = 35\)
⇔ \(5\left( {{x_1}^2 + {x_1}{x_2} + {x_2}^2} \right) = 35\)
⇔ \({x_1}^2 + {x_1}{x_2} + {x_2}^2 = 7\)
⇔ \({\left( {{x_1} - {x_2}} \right)^2} + 3{x_1}{x_2} = 7\)
⇔ \(25 + 3{x_1}{x_2} = 7\)
⇔ \({x_1}{x_2} = - 6\)
⇔ b = −6
Ta có hệ phương trình sau:
\(\left\{ {\begin{array}{*{20}{c}}{{x_1} - {x_2} = 5}\\{{x_1}{x_2} = - 6}\end{array}} \right.\) ⇔ \(\left\{ {\begin{array}{*{20}{c}}{{x_1} = 5 + {x_2}}\\{\left( {5 + {x_2}} \right){x_2} = - 6}\end{array}} \right.\) ⇔ \(\left\{ {\begin{array}{*{20}{c}}{{x_1} = 5 + {x_2}}\\{{x_2}^2 + 5{x_2} + 6 = 0}\end{array}} \right.\)
⇔ \(\left\{ {\begin{array}{*{20}{c}}{{x_1} = 5 + {x_2}}\\{\left( {{x_2} + 2} \right)\left( {{x_2} + 3} \right) = 0}\end{array}} \right.\)
⇔ \(\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{{x_1} = 5 + {x_2}}\\{{x_2} + 2 = 0}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{{x_1} = 5 + {x_2}}\\{{x_2} + 3 = 0}\end{array}} \right.}\end{array}} \right.\) ⇔ \(\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{{x_1} = 3}\\{{x_2} = - 2}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{{x_1} = 2}\\{{x_2} = - 3}\end{array}} \right.}\end{array}} \right.\)
TH1: x1 = 3 và x2 = −2 ⇒ x1 + x2 = 1 = a
⇒ \(\left\{ {\begin{array}{*{20}{c}}{a = 1}\\{b = - 6}\end{array}} \right.\)
TH2: x1 = 2 và x2 = −3 ⇒ x1 + x2 = −1 = a
⇒ \(\left\{ {\begin{array}{*{20}{c}}{a = - 1}\\{b = - 6}\end{array}} \right.\)
Vậy các cặp (a,b) thỏa mãn là (1,−6) hoặc (−1,−6).