Question

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

• A. 1.0
• B. 1
• C. 1.00

Answer 1

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

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

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

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

$\left(iii\right)f\left(0\right)>0\Rightarrow 0^2+2(k+1)0+9k-5>0\Rightarrow 9k-5>0\Rightarrow k>\frac{5}{9}\cdots \left(3\right)$

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


$k\in (\frac{5}{9},1]\cup [6,\infty )$
Hence, the least positve integral value of $k$ is $1$.