Question

For what values of $m$ does the equation $m{x}^2-(m+1)x+2m-1=0$ possess no real roots?

Answer 1

For no real roots, the condition is $D<0$

So, $b^2-4ac<0$

For the equation $m{x}^2-(m+1)x+(2m-1)=0$,

$a=m$ , $b=(m+1)$ and $c=(2m-1)$

So,

$(m+1{)}^2-4\times m\times (2m-1)$ <$0$

$(m^2+1+2m)-8m^2+4m<0$

$-7m^2+6m+1$ < $0$

$7m^2-7m+m-1$ > $0$

$7m(m-1)+1(m-1)>0$

$(7m+1)(m-1)>0$

So, $m<\frac{-1}{7}$ and $m>1$