Question

The range of $k$ for which both the roots of the quadratic equation $(k+1)x^2-3kx+4k=0$ are greater than $1$ is:

• A. $(-\infty ,-1)$
• B. $[-\frac{16}{7},-1]$
• C. $[-\frac{16}{7},-1)$
• D. $(-\frac{16}{7},-1)$

Answer 1

The correct option is C $[-\frac{16}{7},-1)$

Given the quadratic equation:
$(k+1)x^2-3kx+4k=0,k+1\ne 0$
$\Rightarrow x^2-\frac{3kx}{k+1}+\frac{4k}{k+1}=0$
Let $f\left(x\right)=x^2-\frac{3kx}{k+1}+\frac{4k}{k+1}\&\alpha ,\beta$ be the roots of $f\left(x\right)=0.$
Given: $\alpha ,\beta >1$


Conditions:
$\left(i\right)D\ge 0\Rightarrow \frac{9k^2}{(k+1{)}^2}-4\times \frac{4k}{k+1}\ge 0\Rightarrow \frac{9k^2-4(k+1)\left(4k\right)}{(k+1{)}^2}\ge 0\Rightarrow 9k^2-4(k+1)\left(4k\right)\ge 0\Rightarrow 9k^2-16k^2-16k\ge 0\Rightarrow -7k^2-16k\ge 0\Rightarrow k(7k+16)\le 0\Rightarrow k\in [-\frac{16}{7},0]\dots \left(1\right)$

$\left(ii\right)-\frac{b}{2a}>1\text{Here,}a=1,b=\frac{-3k}{k+1}\Rightarrow \frac{3k}{2(k+1)}>1$
$\Rightarrow \frac{k-2}{2(k+1)}>0\Rightarrow k<-1\text{or}k>2\dots \left(2\right)$

$\left(iii\right)f\left(1\right)>0\Rightarrow 1^2-\frac{3k}{k+1}+\frac{4k}{k+1}>0\Rightarrow \frac{2k+1}{k+1}>0$
$\Rightarrow k<-1\text{or}k>-\frac{1}{2}\dots \left(3\right)$

From $\left(1\right),\left(2\right)$ and $\left(3\right)$, we get


or $k\in [-\frac{16}{7},-1)$