Find the nature of the roots of the following quadratic equations.

Solution:

(i) $2 x^{2}-3 x+5=0$

$\mathrm{a}=2, \mathrm{~b}=-3, \mathrm{c}=5$

Discriminant $\mathrm{D}=\mathrm{b}^{2}-4 \mathrm{ac}=9-4 \times 2 \times 5$

$=9-40=-31$

$\Rightarrow \mathrm{D}<0$

Hence, no real root.

(ii) $3 x^{2}-4 \sqrt{3} x+4=0$

$a=3, b=-4 \sqrt{3}, c=4$

Discriminant $\mathrm{D}=\mathrm{b}^{2}-4 \mathrm{ac}$

$=(-4 \sqrt{\mathbf{3}})^{2}-4(3)(4)=48-48=0$

$\Rightarrow D=0$

$\Rightarrow$ Two roots are equal.

The roots are

$=\frac{-\mathbf{b} \pm \sqrt{\mathbf{D}}}{\mathbf{2 a}}=\frac{\mathbf{4} \sqrt{\mathbf{3}} \pm \mathbf{0}}{\mathbf{2} \times \mathbf{3}}=\frac{\mathbf{2}}{\sqrt{\mathbf{3}}}, \frac{\mathbf{2}}{\sqrt{\mathbf{3}}}$

Hence, the roots are $\frac{\mathbf{2}}{\sqrt{\mathbf{3}}}$ and $\frac{\mathbf{2}}{\sqrt{\mathbf{3}}}$.

(iii) $2 x^{2}-6 x+3=0$

$a=2, b=-6, c=3$

Discriminant $D=b^{2}-4 a c=(-6)^{2}-4(2)(3)$

$=36-24=12$

As $D>0$

Therefore, roots are distinct and real.

The roots are

$x=\frac{-\mathbf{b} \pm \sqrt{\mathbf{D}}}{2 \mathbf{a}}$

$=\frac{6 \pm \sqrt{12}}{4}=\frac{6 \pm 2 \sqrt{3}}{4}=\frac{3 \pm \sqrt{3}}{2}$

Therefore, the roots are $\frac{\mathbf{3}+\sqrt{\mathbf{3}}}{\mathbf{2}}$ or $\frac{\mathbf{3}-\sqrt{\mathbf{3}}}{\mathbf{2}}$.