Question

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) \[\sqrt{\dfrac{p}{q}}+\sqrt{\dfrac{q}{p}}-\sqrt{\dfrac{n}{l}}=1\]

Answer 1

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.