If alpha and beta are the zeros of the quadratic polynomial x2+px+q find the polynomial having 1/alpha+1/beta as its zeroes
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I tried but I'm not sure....
x2+px+q
alpha=-b/a beta=q/a
=-p/1 =q/1
1/alpha=-p 1/beta=q
So the polynomial is given by,
k{x2-(1/alpha+1/beta)x+1/alpha*1/beta},k is any constant
=k{x2-(-p+q) x+(-p) (q)}
=x2+px-qx-pq
Hope this helps...
x2+px+q
alpha=-b/a beta=q/a
=-p/1 =q/1
1/alpha=-p 1/beta=q
So the polynomial is given by,
k{x2-(1/alpha+1/beta)x+1/alpha*1/beta},k is any constant
=k{x2-(-p+q) x+(-p) (q)}
=x2+px-qx-pq
Hope this helps...
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