Question

The set of values of $a$ for which one root of the equation $x^2+(2a+1)x+(a-1)=0$ is greater than $1$ and the other root is less than zero is

• A. $a<-\frac{1}{3}$
• B. $0<a<1$
• C. $a>1$
• D. $a\ge 1$

Answer 1

Answer: (A)

$a<-\frac{1}{3}$

Two roots have to exist.
$\Rightarrow \Delta >0=(2a+1{)}^2-4(a-1)>0.$
$4a^2+1+4a-4a+4>0$
$4a^2+5>0$
one root is greater than 1 and other less than 0
$\Rightarrow f\left(1\right)<0andf\left(0\right)<0$
$\Rightarrow 1+2a+1+a-1<0$and $a-1<0soa<1$
$3a+1<0$
$a<\frac{-1}{3}$
$a$ has to be less than $\frac{-1}{3}$