Question

$x^2+(a-b)x+(1-a-b)=0$, where $a,b\in R.$ Find the condition on $a$, for which both roots of the equation are real and unequal.

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

Answer 1

The correct option is C $a>1$

$x^2+(a-b)x+(1-a-b)=0...\text{(i)}$
For given condition,
$D>0$ where, $D=b^2-4ac$

$\Rightarrow (a-b{)}^2-4(1-a-b)>0$
$\Rightarrow b^2-2(a-2)b+(a^2+4a-4)>0$
$\Rightarrow (a-2{)}^2-(a^2+4a-4)<0\Rightarrow -8a+8<0\Rightarrow a>1$