Question 1187915
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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  {{{(a^2 + 1/b^2)*(b^2 + 1/a^2)}}}.



We have

    {{{(a^2 + 1/b^2)*(b^2 + 1/a^2)}}} = {{{a^2*b^2}}} + 1 + 1 + {{{1/(a^2*b^2)}}}.



According to Vieta's theorem,  {{{a*b}}} = {{{-1/2}}}.  THEREFORE,  {{{a^2*b^2}}} = {{{(-1/2)^2}}} = {{{1/4}}}.


HENCE,  
    
    {{{(a^2 + 1/b^2)*(b^2 + 1/a^2)}}} = {{{a^2*b^2}}} + 1 + 1 + {{{1/(a^2*b^2)}}} = {{{1/4 }}}+ 2 + 4 = 6 {{{1/4}}} = 6.25.    <U>ANSWER</U>
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Solved.