Question

Equation $2x^2-2(2a+1)x+a(a+1)=0$ has one root less than $a$ and other root greater than $a$, then which of the following is true

• A. $0<a<1$
• B. $-1<a<0$
• C. $a>0$
• D. $a<-1$

Answer 1

Answer: (A), (C), (D)

$0<a<1$
$a>0$
$a<-1$

Quadratic equation : $2x^2-2(2a+1)x+a(a+1)=0$

Both roots are real, discriminant $=D>0$

Comparing with the standard quadratic equation $A{x}^2+Bx+C=0$, we get $A=2>0$.

Since, $x=a$ lies between the two roots and A is positive, we have $f\left(a\right)<0$

$\Rightarrow$ $2{\left(a\right)}^2-2(2a+1)\left(a\right)+a(a+1)<0$

$\Rightarrow$ $-a^2-a<0$

$\Rightarrow$ $a^2+a>0$

$\Rightarrow$ $a\in (-\infty ,-1)\cup (0,\infty )$

Hence, options (A), (C) and (D) are correct.