Question

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

(i) $k{x}^2+6x+1=0$
(ii) $x^2-kx+9=0$
(iii) $9x^2+3kx+4=0$
(iv) $5x^2-kx+1=0$

Answer 1

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

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

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

$$\begin{gathered} \therefore 36-4k>0 \\ \Rightarrow 4k<36 \\ \Rightarrow k<9 \end{gathered}$$

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

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

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

$$\begin{gathered} \therefore k^2-36>0 \\ \Rightarrow \left(k-6\right)\left(k+6\right)>0 \\ \Rightarrow k<-6\mathrm{or}k>6 \end{gathered}$$

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

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

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

$$\begin{gathered} \therefore 9k^2-144>0 \\ \Rightarrow 9\left(k^2-16\right)>0 \\ \Rightarrow \left(k-4\right)\left(k+4\right)>0 \\ \Rightarrow k<-4\mathrm{or}k>4 \end{gathered}$$

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

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

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

$$\begin{gathered} \therefore k^2-20>0 \\ \Rightarrow k^2-{\left(2\sqrt{5}\right)}^2>0 \\ \Rightarrow \left(k-2\sqrt{5}\right)\left(k+2\sqrt{5}\right)>0 \\ \Rightarrow k<-2\sqrt{5}\mathrm{or}k>2\sqrt{5} \end{gathered}$$