■ No GDC ■Given: \(a + b = 1\) and \({a^4} + {b^4} = 7\)Let \(P = {a^2} + {b^2}\) and \(Q = ab\).\({\left( {a + b} \right)^2} = 1\) and \({\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\); hence, \(P + 2Q = 1\;\;\;\; \Rightarrow \;\;\;\;P = 1 - 2Q\)Also, \({\left( {a + b} \right)^4} = 1\)and \({\left( {a + b} \right)^4} = {a^4} + {b^4} + 4{a^3}b + 4a{b^3} + 6{a^2}{b^2} = \left( {{a^4} + {b^4}} \right) + 4ab\left( {{a^2} + {b^2}} \right) + 6{\left( {ab} \right)^2}\)hence, \(7 + 4PQ + 6{Q^2} = 1\)Substituting \(1 - 2Q\) for \(P\) gives \(7 + 4\left( {1 - 2Q} \right)Q + 6{Q^2} = 7 + 4Q - 2{Q^2} = 1\)Thus, \(7 + 4Q - 2{Q^2} = 1\;\;\;\; \Rightarrow \;\;\;\;2{Q^2} - 4Q - 6 = 0\;\;\;\; \Rightarrow \;\;\;\;{Q^2} - 2Q - 3 = 0\)Solving for \(Q\): \({Q^2} - 2Q - 3 = \left( {Q + 1} \right)\left( {Q - 3} \right) = 0\;\;\;\; \Rightarrow \;\;\;\;Q = - 1\) or \(Q = 3\)But \(Q = 3\) gives \(P = 1 - 2Q = 1 - 2\left( 3 \right) = - 5\) which is not possible because \(P = {a^2} + {b^2} \ge 0\)Thus, \(Q = - 1\) and \(P = 1 - 2\left( { - 1} \right) = 3\)