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

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
$\begin{aligned} & \dfrac{c}{a}+\left| \dfrac{b}{a} \right|=\alpha \beta +\left| \alpha +\beta \right| \\\ & \Rightarrow \dfrac{c}{a}+\left| \dfrac{b}{a} \right|=\left( -1-\lambda \right)\left( 1+\mu \right)+\left| -1-\lambda +1+\mu \right| \\\ & \Rightarrow \dfrac{c}{a}+\left| \dfrac{b}{a} \right|=-1-\mu -\lambda -\lambda \mu +\left| \mu -\lambda \right| \\\ \end{aligned}$
If $\mu >\lambda$ then
$\begin{aligned} & \dfrac{c}{a}+\left| \dfrac{b}{a} \right|=-1-\mu -\lambda -\lambda \mu +\mu -\lambda \\\ & \Rightarrow \dfrac{c}{a}+\left| \dfrac{b}{a} \right|=-1-2\lambda -\lambda \mu \\\ & \Rightarrow \dfrac{c}{a}+\left| \dfrac{b}{a} \right|=-1-\left( 2\lambda +\lambda \mu \right) \\\ \end{aligned}$
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
$\begin{aligned} & \dfrac{c}{a}+\left| \dfrac{b}{a} \right|=-1-\mu -\lambda -\lambda \mu +\lambda -\mu \\\ & \Rightarrow \dfrac{c}{a}+\left| \dfrac{b}{a} \right|=-1-2\mu -\lambda \mu \\\ & \Rightarrow \dfrac{c}{a}+\left| \dfrac{b}{a} \right|=-1-\left( 2\mu +\lambda \mu \right) \\\ \end{aligned}$
For any value of $\mu$ and $\lambda$ the value of $\dfrac{c}{a}+\left| \dfrac{b}{a} \right|$ is always less than $-1$, mathematically
$\therefore \dfrac{c}{a}+\left| \dfrac{b}{a} \right|<-1$