Question

Let a, b, c be real and $a{{x}^{2}}+bx+c=0$ has two real roots, $\alpha $ and $\beta $ where $\alpha <-1$ and $\beta >1$, then show that $\dfrac{c}{a}+\left| \dfrac{b}{a} \right|<-y$ and find $y$.

Answer 1

Hint: We will assume the roots $\alpha $ and $\beta $ as $\alpha +\lambda =-1$ and $\beta =1+\mu $ $\left\{ \lambda ,\mu >0 \right\}$. We have the relation between the roots of the quadratic equation and the coefficients of the quadratic equation i.e. $\alpha +\beta =\dfrac{-b}{a}$ and $\alpha \beta =\dfrac{c}{a}$. From the values of $\alpha $ and $\beta $, known relationships between them we will calculate the value of $\dfrac{c}{a}+\left| \dfrac{b}{a} \right|$. From the obtained value we will conclude the value of $y$.

Complete step-by-step solution
Given that,
Let $a$, $b$, $c$ be real and $a{{x}^{2}}+bx+c=0$ has two real roots, $\alpha $ and $\beta $
Let the values of $\alpha $ and $\beta $ are $\alpha +\lambda =-1$ , $\beta =1+\mu $ $\because \alpha <-1,\beta >1$ where $\left\{ \lambda ,\mu >0 \right\}$.
We have relation between the roots of the quadratic equation and the coefficients of the quadratic equation as
$\alpha +\beta =\dfrac{-b}{a}$ and $\alpha \beta =\dfrac{c}{a}$
Now the value of $\dfrac{c}{a}+\left| \dfrac{b}{a} \right|$ can be calculated as
$\dfrac{c}{a}+\left| \dfrac{b}{a} \right|=\alpha \beta +\left| \alpha +\beta \right|$
The values of $\alpha $ from $\alpha +\lambda =-1$ is $\alpha =-1-\lambda $. Substituting the values of $\alpha $ and $\beta $ in the above equation then we will get
$\begin{align}
  & \dfrac{c}{a}+\left| \dfrac{b}{a} \right|=\alpha \beta +\left| \alpha +\beta \right| \\
 & \Rightarrow \dfrac{c}{a}+\left| \dfrac{b}{a} \right|=\left( -1-\lambda \right)\left( 1+\mu \right)+\left| -1-\lambda +1+\mu \right| \\
 & \Rightarrow \dfrac{c}{a}+\left| \dfrac{b}{a} \right|=-1-\mu -\lambda -\lambda \mu +\left| \mu -\lambda \right| \\
\end{align}$
If $\mu >\lambda $ then
$\begin{align}
  & \dfrac{c}{a}+\left| \dfrac{b}{a} \right|=-1-\mu -\lambda -\lambda \mu +\mu -\lambda \\
 & \Rightarrow \dfrac{c}{a}+\left| \dfrac{b}{a} \right|=-1-2\lambda -\lambda \mu \\
 & \Rightarrow \dfrac{c}{a}+\left| \dfrac{b}{a} \right|=-1-\left( 2\lambda +\lambda \mu \right) \\
\end{align}$
For any value of $\mu $ and $\lambda $ the value of $\dfrac{c}{a}+\left| \dfrac{b}{a} \right|$ is always less than $-1$, mathematically
$\therefore \dfrac{c}{a}+\left| \dfrac{b}{a} \right|<-1$
Now comparing the above expression with the given expression $\dfrac{c}{a}+\left| \dfrac{b}{a} \right|<-y$, then the value of $y$ is equal to $1$.

Note: We can also take the assumption that $\lambda >\mu $, then the value of $\dfrac{c}{a}+\left| \dfrac{b}{a} \right|$ is given by
$\begin{align}
  & \dfrac{c}{a}+\left| \dfrac{b}{a} \right|=-1-\mu -\lambda -\lambda \mu +\lambda -\mu \\
 & \Rightarrow \dfrac{c}{a}+\left| \dfrac{b}{a} \right|=-1-2\mu -\lambda \mu \\
 & \Rightarrow \dfrac{c}{a}+\left| \dfrac{b}{a} \right|=-1-\left( 2\mu +\lambda \mu \right) \\
\end{align}$
For any value of $\mu $ and $\lambda $ the value of $\dfrac{c}{a}+\left| \dfrac{b}{a} \right|$ is always less than $-1$, mathematically
$\therefore \dfrac{c}{a}+\left| \dfrac{b}{a} \right|<-1$