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$p$ and $q$ are positive integers and $\dfrac{p}{q}+\dfrac{q}{p}=3,$ then, $\dfrac{p^{2}}{q^{2}}+\dfrac{q^{2}}{p^{2}}=$

1. $3$
2. $7$
3. $9$
4. $11$

Given that $,\dfrac{p}{q} + \dfrac{q}{p} = 3$

Now, $\left(\dfrac{p}{q} + \dfrac{q}{p} \right)^{2} = 3^{3}$

$\implies \dfrac{p^{2}}{q^{2}} + \dfrac{q^{2}}{p^{2}} + 2 \dfrac{p}{q} \cdot \dfrac{q}{p}= 9 \quad [\because (a+b)^{2} = a^{2} + b^{2} + 2ab]$

$\implies \dfrac{p^{2}}{q^{2}} + \dfrac{q^{2}}{p^{2}} + 2 = 9$

$\implies \dfrac{p^{2}}{q^{2}} + \dfrac{q^{2}}{p^{2}} = 7$

So, the correct answer is $(B).$
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$\frac{p}{q}+\frac{q}{p}=3$

$(\frac{p}{q}+\frac{q}{p})^{2}=3^{2}$

$\frac{p^{2}}{q^{2}}+\frac{q^{2}}{p^{2}}+2(\frac{p}{q})(\frac{q}{p})=9$

$\frac{p^{2}}{q^{2}}+\frac{q^{2}}{p^{2}}+2=9$

$\frac{p^{2}}{q^{2}}+\frac{q^{2}}{p^{2}}=7$
by
180 points