Question

If both the roots of the equation $x^2+2(k+1)x+9k-5=0$ are negative, then the minimum integral value of $k$ is

Answer 1

Let $f\left(x\right)=x^2+2(k+1)x+9k-5$ whose roots are $\alpha ,\beta$
Given: $\alpha ,\beta <0$

Required conditions:
$\left(i\right)D\ge 0\Rightarrow 4(k+1{)}^2-4(9k-5)\ge 0\Rightarrow k^2-7k+6\ge 0$
$\Rightarrow (k-1)(k-6)\ge 0\Rightarrow k\in (-\infty ,1]\cup [6,\infty )\cdots \left(1\right)$

$\left(ii\right)-\frac{b}{a}<0\Rightarrow -2(k+1)<0\Rightarrow k>-1\cdots \left(2\right)$

$\left(iii\right)\frac{c}{a}>0\Rightarrow 9k-5>0\Rightarrow k>\frac{5}{9}\cdots \left(3\right)$

From $\left(1\right),\left(2\right)\text{and}\left(3\right)$,
$k\in (\frac{5}{9},1]\cup [6,\infty )$
Hence, the minimum integral value of $k$ is $1$.