Find the values of k for which the given quadratic equation has real and distinct roots:
Question:

Find the values of k for which the given quadratic equation has real and distinct roots:

(i) $k x^{2}+6 x+1=0$

(ii) $x^{2}-k x+9=0$

(iii) $9 x^{2}+3 k x+4=0$

(iv) $5 x^{2}-k x+1=0$

Solution:

(i) The given equation is $k x^{2}+6 x+1=0$.

$\therefore D=6^{2}-4 \times k \times 1=36-4 k$

The given equation has real and distinct roots if D > 0.

$\therefore 36-4 k>0$

$\Rightarrow 4 k<36$

$\Rightarrow k<9$

(ii) The given equation is $x^{2}-k x+9=0$.

$\therefore D=(-k)^{2}-4 \times 1 \times 9=k^{2}-36$

The given equation has real and distinct roots if D > 0.

$\therefore k^{2}-36>0$

$\Rightarrow(k-6)(k+6)>0$

$\Rightarrow k<-6$ or $k>6$

(iii) The given equation is $9 x^{2}+3 k x+4=0$.

$\therefore D=(3 k)^{2}-4 \times 9 \times 4=9 k^{2}-144$

The given equation has real and distinct roots if D > 0.

$\therefore 9 k^{2}-144>0$

$\Rightarrow 9\left(k^{2}-16\right)>0$

$\Rightarrow(k-4)(k+4)>0$

$\Rightarrow k<-4$ or $k>4$

(iv) The given equation is $5 x^{2}-k x+1=0$.

$\therefore D=(-k)^{2}-4 \times 5 \times 1=k^{2}-20$

The given equation has real and distinct roots if D > 0.

$\therefore k^{2}-20>0$

$\Rightarrow k^{2}-(2 \sqrt{5})^{2}>0$

$\Rightarrow(k-2 \sqrt{5})(k+2 \sqrt{5})>0$

$\Rightarrow k<-2 \sqrt{5}$ or $k>2 \sqrt{5}$