Electronis Discussion

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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 $(B).$

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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