Question

If one root of the equation $\left( l-m \right){{x}^{2}}+lx+1=0$ be double of the other root and if l be real, then $m\le \dfrac{a}{b}$ where a and b are integers. Find min(a,b)

Answer 1

Hint: Assume that the roots of the equation are $\alpha $ and $2\alpha $. Use the fact that the sum of the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ is $\dfrac{-b}{a}$ and the product of roots is $\dfrac{c}{a}$. Hence from two equations in $l,m$ and $\alpha $. One of the equations will be linear in $\alpha $ and the other equation will be quadratic in $\alpha $. Eliminate $\alpha $ from these two equations and hence form a quadratic in l. Use the fact that if the roots of the equation $a{{x}^{2}}+bx+c=0$ are real, then $D={{b}^{2}}-4ac\ge 0$. Hence find the range of values of m over which the above condition is possible. Hence find the value of a and b and hence find a+b.

Complete step-by-step solution -
Let one root of the equation $\left( l-m \right){{x}^{2}}+lx+1=0$ be $\alpha $.
Hence the other root is $2\alpha $.
Now, we know that the sum of the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ is $\dfrac{-b}{a}$
Hence, we have
$\begin{align}
  & \alpha +2\alpha =\dfrac{-l}{l-m} \\
 & \Rightarrow 3\alpha =\dfrac{-l}{l-m} \\
\end{align}$
Dividing both sides by 3, we get
$\alpha =\dfrac{-l}{3\left( l-m \right)}\text{ ……………….. }\left( i \right)$
Also, we know that the product of the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ is $\dfrac{c}{a}$.
Hence, we have
$\begin{align}
  & 2\alpha \times \alpha =\dfrac{1}{l-m} \\
 & \Rightarrow 2{{\alpha }^{2}}=\dfrac{1}{l-m}\text{ ………………. }\left( ii \right) \\
\end{align}$
Substituting the value of $\alpha $ from equation (i) in equation (ii), we get
$\begin{align}
  & 2{{\left( \dfrac{-l}{3\left( l-m \right)} \right)}^{2}}=\dfrac{1}{l-m} \\
 & \Rightarrow \dfrac{2{{l}^{2}}}{9{{\left( l-m \right)}^{2}}}=\dfrac{1}{l-m} \\
\end{align}$
Multiplying both sides by $9{{\left( l-m \right)}^{2}}$, we get
$2{{l}^{2}}=9\left( l-m \right)$
Adding $9m-9l$ on both sides of the equation, we get
$2{{l}^{2}}-9l+9m=0\text{ }\left( a \right)$, which is a quadratic in l.
We know that if the roots of the equation $a{{x}^{2}}+bx+c=0$ are real, then $D={{b}^{2}}-4ac\ge 0$.
Since l is real, we have
${{9}^{2}}-4\left( 9m \right)\left( 2 \right)\ge 0$
Adding 72m on both sides of the equation, we get
$72m\le 81$
Dividing both sides by 72, we get
$m\le \dfrac{81}{72}=\dfrac{9}{8}$
Hence, we have
$m\le \dfrac{9}{8}$
Hence, we have a= 9 and b = 8
Hence a+b = 17
Hence Min(a+b) = 17.

Note: [1] In the above question we have used that the minimum of sum of numerator and denominator of a fraction in $\dfrac{p}{q},p,q\in \mathbb{Z},q\ne 0$ form is when the fraction is expressed in lowest terms.
[2] We can also directly arrive at the above condition (a) using quadratic formula. Since one of the roots is twice the other, we have $\dfrac{-l+\sqrt{{{l}^{2}}-4\left( l-m \right)}}{2\left( l-m \right)}=2\dfrac{-l-\sqrt{{{l}^{2}}-4\left( l-m \right)}}{2\left( l-m \right)}$
Hence, we have
$\begin{align}
  & -l+\sqrt{{{l}^{2}}-4\left( l-m \right)}=-2l-2\sqrt{{{l}^{2}}-4\left( l-m \right)} \\
 & \Rightarrow 3\sqrt{{{l}^{2}}-4\left( l-m \right)}=-l \\
\end{align}$
Squaring both sides, we get
$9{{l}^{2}}-36l+36m={{l}^{2}}\Rightarrow 8{{l}^{2}}-36l+36m=0$
Dividing both sides by 4, we get
$2{{l}^{2}}-9l+9m=0$, which is the same as equation (a).
Proceeding similarly as above we get $m\le \dfrac{9}{8}$