6p² +17p + 12 = 0 6p² + 9p + 8p + 12 = 0 ⟹ 3 p (2p + 3) + 4 (2p + 3) = 0 ⟹ (3p + 4) (2p + 3) = 0 ⟹ p = - 4/3, - 3/2 II. 12q² - 25q + 7 = 0 ⟹ 12 q² - 21q - 4q + 7 =0 ⟹ 3q(4q - 7) - 1(4q -7)= 0 ⟹ (3q - 1) (4q - 7)= 0 ⟹ q =1/3 , 7/4 Hence p < q.
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