Question

Find the values of $k$ for which the given quadratic equation has real and distinct roots:
$\left(i\right)k{x}^2+2x+1=0\text{(ii)}k{x}^2+6x+1=0$

Answer 1

$i)k{x}^2+2x+1$

$=b^2-4ac>0$

$=(2{)}^2-4(k\left)\right(1)>0$

$=4k<4\Rightarrow k<1$

$ii)k{x}^2+6x+1-0$

$=(6{)}^2-4(k\left)\right(1)>0$

$=4k<36$

$\Rightarrow k<9$