Of one zero of 2x2-5x+k is the resiprocal of the other ,then find the other,find the value of k
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Let the zero's be a and 1/a.Then ,
(x-a)(ax-1) = ax^2-x-a^2x+a
= ax^2-(1+a^2)x+a
On putting a = 2 , we get :
2x^2 - 5x+2
Therefore, k = 2
(x-a)(ax-1) = ax^2-x-a^2x+a
= ax^2-(1+a^2)x+a
On putting a = 2 , we get :
2x^2 - 5x+2
Therefore, k = 2
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