Find the nature of roots of the following quadratic equations :

(i) $x^2 - \dfrac{1}{2}x - \dfrac{1}{2} = 0$

(ii)$x^2 - 2\sqrt{3}x - 1 = 0$

If real roots exist, find them.

(i) The given equation is $x^2 - \dfrac{1}{2}x - \dfrac{1}{2} = 0$

Comparing it with ax² + bx + c = 0, we get
a = 1 , b = $-\dfrac{1}{2}$ , c = $-\dfrac{1}{2}$

∴ Discriminant = b² - 4ac

Putting values of a, b, c in formula

$\Big(-\dfrac{1}{2}\Big)^2 - 4 \times 1 \times -\dfrac{1}{2} \\[1em] = \dfrac{1}{4} + 2 \\[1em] = \dfrac{1 + 8}{4} \\[1em] = \dfrac{9}{4} \\[1em]$

Discriminant = $\dfrac{9}{4}$
Since Discriminant > 0, hence the given equation has two distinct real roots.

The roots of the equation are given by:

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a} \\[1em] \Rightarrow x = \dfrac{-\Big(-\dfrac{1}{2}\Big) ± \sqrt{\Big(-\dfrac{1}{2}\Big)^2 - 4\times 1 \times -\dfrac{1}{2}}}{2 \times 1} \\[1em] \Rightarrow x = \dfrac{\dfrac{1}{2} ± \sqrt{\dfrac{1}{4} + 2}}{2} \\[1em] \Rightarrow x = \dfrac{\dfrac{1}{2} ± \sqrt{\dfrac{9}{4}}}{2} \\[1em] \Rightarrow x = \dfrac{\dfrac{1}{2} + \sqrt{\dfrac{9}{4}}}{2} \text{ or } \dfrac{\dfrac{1}{2} - \sqrt{\dfrac{9}{4}}}{2} \\[1em] \Rightarrow x = \dfrac{\dfrac{1}{2} + \dfrac{3}{2}}{2} \text{ or } \dfrac{\dfrac{1}{2} - \dfrac{3}{2}}{2} \\[1em] \Rightarrow x = \dfrac{\dfrac{4}{2}}{2} \text{ or } \dfrac{-\dfrac{2}{2}}{2} \\[1em] \Rightarrow x = \dfrac{4}{4} \text{ or } -\dfrac{2}{4} \\[1em] \Rightarrow x = 1 \text{ or } -\dfrac{1}{2}$

Hence roots of the given equations are 1 , $-\dfrac{1}{2}$.

(ii) The given equation is $x^2 - 2\sqrt{3}x - 1 = 0$
Comparing it with ax² + bx + c = 0, we get
a = 1 , b = $-2\sqrt{3}$ , c = -1
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$(-2\sqrt{3})^2 - 4 \times 1 \times -1 \\[1em] = 12 + 4 \\[1em] = 16 \\[1em]$

Discriminant = 16 , Since Discriminant > 0, hence given equation have real and distinct roots.

The roots of the equation are given by

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a} \\[1em] \Rightarrow x = \dfrac{-(-2\sqrt{3}) ± \sqrt{(-2\sqrt{3})^2 - 4\times 1 \times -1}}{2 \times 1} \\[1em] \Rightarrow x = \dfrac{2\sqrt{3} ± \sqrt{12 + 4}}{2} \\[1em] \Rightarrow x = \dfrac{2\sqrt{3} ± \sqrt{16}}{2} \\[1em] \Rightarrow x = \dfrac{2\sqrt{3} + 4}{2} \text{ or } \dfrac{2\sqrt{3} - 4}{2} \\[1em] \Rightarrow x = \sqrt{3} + 2 \text{ or } \sqrt{3} - 2 \\[1em]$

Hence roots of the given equations are , $\sqrt{3} + 2 , \sqrt{3} - 2$ .