Let and be the roots of the quadratic equation x^2 + bx + c = 0 , where b > 0 and c > 0 . If ^2 + ^2 = 7 and ^4 + ^4 = 47 , then the quadratic equation whose roots are ^3 and ^3 is :

Step-by-step solution

Given the quadratic equation x^2 + bx + c = 0 , the sum of roots is + = -b and the product of roots is = c . We are given ^2 + ^2 = 7 and ^4 + ^4 = 47 . Using the identity ^4 + ^4 = ( ^2 + ^2)^2 - 2( )^2 , we have: 47 = (7)^2 - 2c^2 47 = 49 - 2c^2 2c^2 = 2 c^2 = 1 Since c > 0 , we get c = 1 . Now, using the identity ^2 + ^2 = ( + )^2 - 2 , we have: 7 = (-b)^2 - 2(1) b^2 = 9 Since b > 0 , we get b = 3 . Thus, the base equation is x^2 + 3x + 1 = 0 , which gives + = -3 and = 1 . We need the quadratic equation whose ro