Question

If atleast one of the root of the equation $x^2-(a+2)x+\frac{7a}{4}=0$ is less than $0$, then $a$ 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 $f\left(x\right)=x^2-(a+2)x+\frac{7a}{4}$

Case $1:$ Both the roots are negative
$\left(i\right)D\ge 0$
$D=(a+2{)}^2-7a\ge 0$
$\Rightarrow a^2-3a+4\ge 0\Rightarrow {(a-\frac{3}{2})}^2+\frac{15}{4}\ge 0\Rightarrow a\in R$

$\left(ii\right)f\left(0\right)>0\Rightarrow a>0$

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

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

Case $2:$ One root is positive and other root is negative
So, $0$ lies in between the roots.
$\Rightarrow f\left(0\right)<0$
$\Rightarrow a<0\cdots \left(2\right)$

Case $3:$ One roots is $0$ and other is negative.
$\left(i\right)f\left(0\right)=0\Rightarrow a=0$
$\left(ii\right)-\frac{b}{2a}<0\Rightarrow a<-2$
$\therefore a\in \varphi \cdots \left(3\right)$

From $\left(1\right),\left(2\right)$ and $\left(3\right)$
$a\in (-\infty ,0)$