Question

If at least one of the roots of the equation $x^2-(k+2)x+\frac{7k}{4}=0$ is negative, then $k$ lies in the interval

• A. $(-\infty ,1]\cup [4,\infty )$
• B. $R-\left\{0\right\}$
• C. $(-\infty ,0)\cup [4,\infty )$
• D. $(-\infty ,0)$

Answer 1

The correct option is D $(-\infty ,0)$

Let $f\left(x\right)=x^2-(k+2)x+\frac{7k}{4}$

Case $1:$ Both the roots are negative
$\left(i\right)D\ge 0$
$D=(k+2{)}^2-7k\ge 0$
$\Rightarrow k^2-3k+4\ge 0\Rightarrow {(k-\frac{3}{2})}^2+\frac{7}{4}\ge 0\Rightarrow k\in R$

$\left(ii\right)\frac{c}{a}>0\Rightarrow k>0$

$\left(iii\right)-\frac{b}{a}<0\Rightarrow k+2<0\Rightarrow k<-2$

$\therefore k\in \varphi \cdots \left(1\right)$

Case $2:$ One root is positive and other root is negative
$\Rightarrow \frac{c}{a}<0$
$\Rightarrow k<0\cdots \left(2\right)$

Checking boundary condition for case $:2$
For $k=0,$ we get
$x^2-2x=0$
$\Rightarrow x=0,2$
Here, no root is negative.

$\therefore k\in \left(1\right)\cup \left(2\right)$
$\Rightarrow k\in \varphi \cup (-\infty ,0)$
$\Rightarrow k\in (-\infty ,0)$