Question

Explain the nature of roots of a quadratic equation $x^2-x-2=0$.

Answer 1

Given that,

$x^2-x-2=0$

Comparing with $a{x}^2+bx+c=0$, we get,

$a=1,b=-1\text{and}c=-2$

We know that, $D=b^2-4ac$

$\therefore D=(-1{)}^2-4(1\left)\right(-2)=1+8=9$

$\Rightarrow D>0$

Since, the discriminant is positive, the given equation will have two distinct real roots.