Question

For what values of K the quadratic equation
$a^2-Ka+1=0$ has equal roots

Answer 1

Here, $a^2-ka+1=0$

which is of the form $\alpha x^2+\beta x+\gamma =0$

$\therefore \alpha =1,\beta =-k,\gamma =1$

now, Discriminant, $D={\beta }^2-4\alpha \gamma$

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

$=k^2-4$

Since the roots are equal , D=0

$\Rightarrow k^2-4=0$

$\Rightarrow k^2=4$

$\Rightarrow k=\pm 2$

$\therefore k=2$ or $k=-2$