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ATQ, 2p²+12p+18=0 2p²+6p+6p+18=0 2p(p+3) +6(p+3) =0 (2p+6)(p+3)=0 p =-3,-3 ll) 3q²+13q+12=0 3q²+9q+4q+12=0 3q(q+3) +4(q+3) =0 (3q+4)(q+3)=0 q = - 4/3, -3 Hence, p ≤ q
The initial codons are
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