Instructions

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

Question 119

I. $$p^{2}+13p+42=0$$ II. $$q^{2}=36$$

Solution

$$p^2+13p+42 = 0$$
$$(p+6)(p+7) = 0$$
$$p = -6, -7$$

$$q^2 = 36$$
$$q = -6, 6$$

$$p\leq q$$


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