Instructions

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

Question 118

I. $$p=\frac{\sqrt{4}}{\sqrt{9}}$$ II. $$9q^{2}-12q+4=0$$

Solution

$$p = \frac{\sqrt{4}}{\sqrt{9}}$$
$$p = \frac{2}{3}$$

$$9q^2-12q+4 = 0$$
$${(3q-2)}^2 = 0$$
$$q = \frac{2}{3}$$

p = q


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