Find the values of k for each of the following quadratic equations,
Question.

Find the values of k for each of the following quadratic equations, so that they have two real equal roots.

(i) $2 x^{2}+k x+3=0$

(ii) $k x(x-2)+6=0$


Solution:

(i) $2 x^{2}+k x+3=0$

$\mathrm{a}=2, \mathrm{~b}=\mathrm{k}, \mathrm{c}=3$

$\mathrm{D}=\mathrm{b}^{2}-4 \mathrm{ac}=\mathrm{k}^{2}-4 \times 2 \times 3=\mathrm{k}^{2}-24$

Two roots will be equal

if $D=0$, i.e., if $k^{2}-24=0$

i.e., if $k^{2}-24$, i.e., if $k=\pm \sqrt{\mathbf{2 4}}$

i.e., if $\mathrm{k}=\pm \mathbf{2} \sqrt{\mathbf{6}}$

(ii) $\quad \mathrm{kx}(\mathrm{x}-2)+6=0$

or $k x^{2}-2 k x+6=0$

$\mathrm{a}=\mathrm{k}, \mathrm{b}=-2 \mathrm{k}, \mathrm{c}=6$

Discriminant $D=b^{2}-4 a c=(-2 k)^{2}-4(k)(6)$

$=4 k^{2}-24 k$

Two roots will be equal

if $D=0$

$4 k^{2}-24 k=0$

$4 k(k-6)=0$

Either $4 \mathrm{k}=0$ or $\mathrm{k}-6=0$

$\mathrm{k}=0$ or $\mathrm{k}=6$

However, if $k=0$, then the equation will not have the terms ‘ $x^{2}$ ‘ and ‘ $x$ ‘.

Hence $\mathrm{k}=6$
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