Answer

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**Hint:**We solve this problem by using the definition of ratios that is if the ratio of two numbers is \[a:b\] then there exist a number \['k'\] such that the numbers are \[ak,bk\] then we use the sum and product of roots of quadratic equation that is if the equation \[a{{x}^{2}}+bx+c=0\] have the roots \[\alpha ,\beta \] then the sum of roots is

\[\alpha +\beta =\dfrac{-b}{a}\]

Also the product of roots is given as

\[\alpha \beta =\dfrac{c}{a}\]

By using the above formulas we find the required result.

**Complete step-by-step answer:**

We are given that the quadratic equation as

\[l{{x}^{2}}+nx+n=0\]

We are given that the roots of this equation are in the ratio \[p:q\]

We now that if the ratio of two numbers is \[a:b\] then there exist a number \['k'\] such that the numbers are \[ak,bk\]

By using this result we get the roots of given equation as \[pk,qk\]

We know that, if the equation \[a{{x}^{2}}+bx+c=0\] has the roots \[\alpha ,\beta \] then the sum of roots is

\[\alpha +\beta =\dfrac{-b}{a}\]

Also the product of roots is given as

\[\alpha \beta =\dfrac{c}{a}\]

Now, by using the product of roots of given equation we get

\[\begin{align}

& \Rightarrow \left( pk \right)\left( qk \right)=\dfrac{n}{l} \\

& \Rightarrow {{k}^{2}}=\left( \dfrac{1}{pq} \right)\left( \dfrac{n}{l} \right) \\

& \Rightarrow k=\pm \sqrt{\dfrac{1}{pq}}\sqrt{\dfrac{n}{l}} \\

\end{align}\]

Now, by using the sum of roots we get

\[\Rightarrow pk+qk=\dfrac{-n}{l}\]

Now, by substituting the value of \['k'\] in above equation we get

\[\begin{align}

& \Rightarrow p\left( \pm \sqrt{\dfrac{1}{pq}}\sqrt{\dfrac{n}{l}} \right)+q\left( \pm \sqrt{\dfrac{1}{pq}}\sqrt{\dfrac{n}{l}} \right)=\dfrac{-n}{l} \\

& \Rightarrow p\left( \pm \sqrt{\dfrac{1}{pq}}\sqrt{\dfrac{n}{l}} \right)+q\left( \pm \sqrt{\dfrac{1}{pq}}\sqrt{\dfrac{n}{l}} \right)+{{\left( \pm \sqrt{\dfrac{n}{l}} \right)}^{2}}=0 \\

\end{align}\]

Now by taking the common term out we get

\[\Rightarrow \left( \pm \sqrt{\dfrac{n}{l}} \right)\left( \dfrac{p}{\sqrt{pq}}+\dfrac{q}{\sqrt{pq}}\pm \sqrt{\dfrac{n}{l}} \right)=0\]

We know that for any number we have

\[\Rightarrow a=\sqrt{{{a}^{2}}}\]

By using this formula to above equation we get

\[\begin{align}

& \Rightarrow \sqrt{\dfrac{{{p}^{2}}}{pq}}+\sqrt{\dfrac{{{q}^{2}}}{pq}}\pm \sqrt{\dfrac{n}{l}}=0 \\

& \Rightarrow \sqrt{\dfrac{p}{q}}+\sqrt{\dfrac{q}{p}}\pm \sqrt{\dfrac{n}{l}}=0 \\

\end{align}\]

Here we can see that there will be two equations one for positive sign and other for negative sign.

By taking the positive sign we get

\[\Rightarrow \sqrt{\dfrac{p}{q}}+\sqrt{\dfrac{q}{p}}+\sqrt{\dfrac{n}{l}}=0\]

Now, by taking the negative sign we get

\[\Rightarrow \sqrt{\dfrac{p}{q}}+\sqrt{\dfrac{q}{p}}-\sqrt{\dfrac{n}{l}}=0\]

Here, we can see that both the equations may be correct.

Now, by comparing the above equations we got with the given options we get

\[\therefore \sqrt{\dfrac{p}{q}}+\sqrt{\dfrac{q}{p}}+\sqrt{\dfrac{n}{l}}=0\]

**So, the correct answer is “Option (b)”.**

**Note:**Students may do mistake in taking the root value.

Here we have the value of \['k'\] as

\[\Rightarrow k=\pm \sqrt{\dfrac{1}{pq}}\sqrt{\dfrac{n}{l}}\]

Here students may miss the \['\pm '\] sign which results to decrease of one possible equation.

We get the possible equation as

\[\Rightarrow \sqrt{\dfrac{p}{q}}+\sqrt{\dfrac{q}{p}}-\sqrt{\dfrac{n}{l}}=0\]

\[\Rightarrow \sqrt{\dfrac{p}{q}}+\sqrt{\dfrac{q}{p}}+\sqrt{\dfrac{n}{l}}=0\]

But, if we do not consider we miss one possibility which is also correct relation.

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