Question

Determine the set of values of $k$ for which the given quadratic equation has real roots:
$x^2-kx+9=0$

Answer 1

We have,

$x^2-kx+a=0$

comparing with $a{x}^2+bx+c=0$

$\therefore a=1,b=k,c=9$

$\Delta =b^2-4ac$

$=(k{)}^2-4\times 1\times 9$

=k^2-36$

$\therefore \Delta =0$

$k^2=36$

$k^2=6^2$

$k=6$

Hence, this is answer.