Question

If both the roots of $x^2+2(k+1)x+9k-5=0$ are less than $0$ and $k<100,$ then the number of integral values of $k$ is

• A. $92$
• B. $93$
• C. $94$
• D. $95$

Answer 1

The correct option is D $95$

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

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

$\left(ii\right)\alpha +\beta <0\Rightarrow -2(k+1)<0\Rightarrow k>-1\cdots \left(2\right)$

$\left(iii\right)\alpha \cdot \beta >0\Rightarrow 9k-5>0\Rightarrow k>\frac{5}{9}\cdots \left(3\right)$

From $\left(1\right),\left(2\right)\text{and}\left(3\right)$,
$k\in [6,\infty )\cup \left\{1\right\}$
Since, $k$ is an integer less than $100$,
$k=1,6,7,\dots ,98,99$