SOLUTION: the roots of 2x^2 +3x-1=0 are alpha and beta. Find the values of [alpha square +1÷beta square] [beta square+1÷alpha square]

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: the roots of 2x^2 +3x-1=0 are alpha and beta. Find the values of [alpha square +1÷beta square] [beta square+1÷alpha square]       Log On


   

Question 1187915: the roots of 2x^2 +3x-1=0 are alpha and beta. Find the values of [alpha square +1÷beta square] [beta square+1÷alpha square]

Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
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the roots of 2x^2 +3x-1=0 are alpha and beta.
Find the values of [alpha square +1÷beta square] [beta square+1÷alpha square]
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To make my writing easier, I will use  "a" instead of alpha  and "b" instead of bete.


So, I need find  %28a%5E2+%2B+1%2Fb%5E2%29%2A%28b%5E2+%2B+1%2Fa%5E2%29.



We have

    %28a%5E2+%2B+1%2Fb%5E2%29%2A%28b%5E2+%2B+1%2Fa%5E2%29 = a%5E2%2Ab%5E2 + 1 + 1 + 1%2F%28a%5E2%2Ab%5E2%29.



According to Vieta's theorem,  a%2Ab = -1%2F2.  THEREFORE,  a%5E2%2Ab%5E2 = %28-1%2F2%29%5E2 = 1%2F4.


HENCE,  
    
    %28a%5E2+%2B+1%2Fb%5E2%29%2A%28b%5E2+%2B+1%2Fa%5E2%29 = a%5E2%2Ab%5E2 + 1 + 1 + 1%2F%28a%5E2%2Ab%5E2%29 = 1%2F4++ 2 + 4 = 6 1%2F4 = 6.25.    ANSWER

Solved.