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# The ratio of roots of equation $l{{x}^{2}}+nx+n=0$ is $p:q$ then(a) $\sqrt{\dfrac{q}{p}}+\sqrt{\dfrac{p}{q}}-\sqrt{\dfrac{l}{n}}=-1$(b) $\sqrt{\dfrac{p}{q}}+\sqrt{\dfrac{q}{p}}+\sqrt{\dfrac{n}{l}}=0$(c) $\sqrt{\dfrac{q}{p}}-\sqrt{\dfrac{p}{q}}+\sqrt{\dfrac{l}{n}}=0$(d) $\sqrt{\dfrac{p}{q}}+\sqrt{\dfrac{q}{p}}-\sqrt{\dfrac{n}{l}}=1$

Last updated date: 10th Aug 2024
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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.

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.