GATE ECE 2021 | GA Question: 2

Question

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

• A. $3$
• B. $7$
• C. $9$
• D. $11$

Answer 1

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 $(B).$

Answer 2

$\frac{p}{q}+\frac{q}{p}=3$

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

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

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

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