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# Help!

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Given two real numbers 1 $$\frac{1}{p} + \frac{1}{q} = 1$$and $$pq = \frac{9}{2}$$, what is q?

Jul 8, 2020

#1
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Notice that

$$\dfrac1p = \dfrac{q}{pq}\\ \dfrac1q = \dfrac{p}{pq}$$

If we rewrote the fractions like this, then we could add the two fractions.

$$\dfrac{1}p + \dfrac1q = \dfrac{q}{pq} + \dfrac{p}{pq} = \dfrac{p + q}{pq}$$

Substituting pq = 9/2,

$$p + q = \dfrac92$$

Subtracting q on both sides,

$$p = \dfrac92 - q$$

Substituting this into pq = 9/2,

$$q\left(\dfrac92 - q\right) = \dfrac92\\ q^2 - \dfrac92 q + \dfrac92 = 0\\ 2q^2 - 9q + 9 = 0$$

You can solve this equation by using the formula $$q = \dfrac{-b\pm\sqrt{b^2- 4ac}}{2a}$$, where a is the coefficient of x^2, b is the coefficient of x, and c is the constant term.

Jul 8, 2020
#2
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Thanks!

AnimalMaster  Jul 8, 2020