Instructions

For the two given equations I and II----

Question 116

I. $$3p^{2}+2p-1=0$$ II. $$2q^{2}+7q+6=0$$

Solution

$$3p^2+2p-1 = 0$$
$$(3p-1)(p+1) = 0$$
$$p = -1, \frac{1}{3}$$

$$2q^2+7q+6 = 0$$
$$(2q+3)(q+2) = 0$$
$$q = -2, -\frac{3}{2}$$

p > q


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