Question

Determine the set of values of $k$ for which the given quadratic equation has real roots:
$4x^2-3kx+1=0$

Answer 1

We have,

$4x^2-3kx+1=0$

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

$\therefore a=4,b=3k,c=1$

$\Delta =b^2-4ac$

$=(3k{)}^2-4\times 4\times 1$

$=9k^2-16$

$\therefore \Delta =0$

$k^2=\frac{16}{9}$

$\therefore k=\pm \frac{4}{3}$

Hence, this is answer.