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  • Suppose alpha , beta are roots of x^2-7x+8=0, with alpha>beta , then value of 16/alpha +3 (beta)^2-19 beta is

Suppose alpha , beta are roots of x^2-7x+8=0, with alpha>beta , then value of 16/alpha +3 (beta)^2-19 beta is

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Solution:     eta ^2=7eta -8.

               Thus ,    16/alpha +3eta ^2-19eta =16/alpha +3(7eta -8)-19eta

               Rightarrow          = = 16/alpha +2eta -24

              Also ,   alpha =7+sqrt17/2  ,  eta =7-sqrt17/2

            herefore   1/alpha =2 / 7+sqrt17=2(7-sqrt17)/49-17=7-sqrt17/16

        Thus ,   16/alpha +3eta ^2-19eta =(7-sqrt17)+(7-sqrt17)-24=-10 

Posted by

Deependra Verma

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