Let and be the distinct roots of the equation (x^ ₃ 5 )^2 - 26 5^ ₃ x + 25 = 0 . The quadratic equation whose roots are +1 and +1 is

Step-by-step solution

Given the equation (x^ ₃ 5 )^2 - 26 5^ ₃ x + 25 = 0 . Using the logarithmic identity a^ _b c = c^ _b a , we can rewrite x^ ₃ 5 as 5^ ₃ x . Let y = 5^ ₃ x . The equation becomes: y^2 - 26y + 25 = 0 (y - 25)(y - 1) = 0 So, y = 25 or y = 1 . Case 1: 5^ ₃ x = 25 5^ ₃ x = 5^2 ₃ x = 2 x = 3^2 = 9 . Case 2: 5^ ₃ x = 1 5^ ₃ x = 5^0 ₃ x = 0 x = 3^0 = 1 . The distinct roots are = 9 and = 1 . We need to find the quadratic equation whose roots are +1 and +1 . Substituting the values of and : First root = 9 1+1 = 9 2 Second roo