Question

If the ratio of the roots of the equation $ {{x}^{2}}+qx+p=0 $ be equal to the ratio of the roots of the equation $ {{x}^{2}}+lx+m=0 $ , find the relation of $ p,q,l,m $ ?

Answer 1

Hint: We first take the general formula of roots for quadratic equations. We apply the roots for the equations $ {{x}^{2}}+qx+p=0 $ and $ {{x}^{2}}+lx+m=0 $ . Then we use the relation of the ratio of roots being equal to find the relation of $ p,q,l,m $ .

Complete step by step solution:
We know that for quadratic equation $ a{{x}^{2}}+bx+c=0 $ , if the roots are $ m,n $ then we can say that
 $ m+n=-\dfrac{b}{a} $ and $ mn=\dfrac{c}{a} $ .
Let us assume the roots of the equation $ {{x}^{2}}+qx+p=0 $ to be $ a,b $ .
Therefore, $ a+b=-q $ and $ ab=p $ .
We also assume the roots of the equation $ {{x}^{2}}+lx+m=0 $ to be $ c,d $ .
Therefore, $ c+d=-l $ and $ cd=m $ .
It is also given that the ratio of the roots of the equation $ {{x}^{2}}+qx+p=0 $ be equal to the ratio of the roots of the equation $ {{x}^{2}}+lx+m=0 $ .
Therefore, $ \dfrac{a}{b}=\dfrac{c}{d} $ .
Using componendo-dividendo for $ \dfrac{a}{b}=\dfrac{c}{d} $ we get $ \dfrac{a+b}{a-b}=\dfrac{c+d}{c-d} $ .
We take square on both sides and get $ \dfrac{{{\left( a+b \right)}^{2}}}{{{\left( a-b \right)}^{2}}}=\dfrac{{{\left( c+d \right)}^{2}}}{{{\left( c-d \right)}^{2}}} $ .
We put the values in numerators and simplify the denominator part to get
\[\begin{align}
  & \dfrac{{{\left( a+b \right)}^{2}}}{{{\left( a-b \right)}^{2}}}=\dfrac{{{\left( c+d \right)}^{2}}}{{{\left( c-d \right)}^{2}}} \\
 & \Rightarrow \dfrac{{{\left( -q \right)}^{2}}}{{{\left( a+b \right)}^{2}}-4ab}=\dfrac{{{\left( -l \right)}^{2}}}{{{\left( c+d \right)}^{2}}-4cd} \\
\end{align}\]
Again, we put the values to get
\[\begin{align}
  & \dfrac{{{\left( -q \right)}^{2}}}{{{\left( a+b \right)}^{2}}-4ab}=\dfrac{{{\left( -l \right)}^{2}}}{{{\left( c+d \right)}^{2}}-4cd} \\
 & \Rightarrow \dfrac{{{q}^{2}}}{{{q}^{2}}-4p}=\dfrac{{{l}^{2}}}{{{l}^{2}}-4m} \\
\end{align}\]
Now we complete cross multiplication to get
\[\begin{align}
  & \dfrac{{{q}^{2}}}{{{q}^{2}}-4p}=\dfrac{{{l}^{2}}}{{{l}^{2}}-4m} \\
 & \Rightarrow {{q}^{2}}{{l}^{2}}-4m{{q}^{2}}={{q}^{2}}{{l}^{2}}-4p{{l}^{2}} \\
 & \Rightarrow p{{l}^{2}}=m{{q}^{2}} \\
\end{align}\]
Therefore, the relation of $ p,q,l,m $ is \[p{{l}^{2}}=m{{q}^{2}}\].
So, the correct answer is “\[p{{l}^{2}}=m{{q}^{2}}\]”.

Note: The componendo-dividendo formula can be used as the inverse form where we take $ \dfrac{a}{b}=\dfrac{c}{d}\Rightarrow \dfrac{a-b}{a+b}=\dfrac{c-d}{c+d} $ . We put the values and the relation still remains the same for \[p{{l}^{2}}=m{{q}^{2}}\].