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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
avatar+8352 
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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
avatar+794 
0

Thanks!

AnimalMaster  Jul 8, 2020

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