Question

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

• A. 95.0
• B. 95
• C. 95.00

Answer 1

Answer: (A), (B), (C)

Let $f\left(x\right)=x^2+2(k+1)x+9k-5$ and roots of $f\left(x\right)=0$ are $\alpha ,\beta$ such that $\alpha ,\beta <0$

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

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

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

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


$k\in (\frac{5}{9},1]\cup [6,\infty )$
Since $k$ is an integer less than $100,$
$k=1,6,7,\dots ,98,99$
$\Rightarrow 95$ possible values of $k$