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f(x)=ax2−(3+2a)x+6,a≠0\large f(x)=ax^2 - (3 + 2a)x + 6, \quad a\ne 0f(x)=ax2−(3+2a)x+6,a=0
If the set of values of aaa for which f(x)f(x)f(x) is postive for exactly three distinct negative integral values of xxx is (c,d](c,d](c,d], then find the value of c2+16d2c^2 + 16d^2c2+16d2.
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