If alpha+beeta are the zeros of 7x^2-19x-6,Find
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7x²-19x-6
7x²-21x+2x-6
7x(x-3)+2(x-3)
7x+2,x-3
7x+2=0,x-3=0
7x=-2,x=3
x=-2/7,x=3
(alpha²+beta²)=(alpha+beta)²-2alpha *beta. =(-2)²-2*-2/7
=4 -14-2/7
=4-16/7
=-12/7
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