Discuss the nature of the roots of the following equations :

(i) $3x^2 - 7x + 8 = 0$

(ii) $x^2 - \dfrac{1}{2}x - 4 = 0$

(iii) $5x^2 - 6\sqrt{5}x + 9 = 0$

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

In case real roots exist , then find them.

(i) The given equation is 3x² - 7x + 8 = 0

Comparing it with ax² + bx + c = 0
a= 3, b = -7, c = 8

$\therefore \text{Discriminant }= b^2 - 4ac \\[0.5em] = (-7)^2 - 4 \times 3 \times 8 \\[0.5em] = 49 - 96 \\[0.5em] = -47 \lt 0$

Since, Discriminant < 0 , hence equation has no real roots.

(ii) The given equation is $x^2 - \dfrac{1}{2}x - 4 = 0$

Comparing it with ax² + bx + c = 0
a= 1, b = $-\dfrac{1}{2}$, c = -4

$\therefore \text{Discriminant } = b^2 - 4ac \\[1em] = (-\dfrac{1}{2})^2 - 4 \times 1 \times -4 \\[1em] = \dfrac{1}{4} + 16 \\[1em] = \dfrac{65}{4} \gt 0$

Since, Discriminant > 0 , hence equation has two distinct and real roots.

By using the formula , $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$\Rightarrow \dfrac{-(-\dfrac{1}{2}) ± \sqrt{(-\dfrac{1}{2})^2 - 4 \times 1 \times -4}}{2 \times 1} \\[1em] \Rightarrow \dfrac{\dfrac{1}{2} ± \sqrt{\dfrac{1}{4} + 16}}{2} \\[1em] \Rightarrow \dfrac{\dfrac{1}{2} + \sqrt{\dfrac{65}{4}}}{2} \text{ or } \dfrac{\dfrac{1}{2} - \sqrt{\dfrac{65}{4}}}{2} \\[1em] \Rightarrow \dfrac{\dfrac{1}{2} + \dfrac{\sqrt{65}}{2}}{2} \text{ or } \dfrac{\dfrac{1}{2} - \dfrac{\sqrt{65}}{2}}{2} \\[1em] \Rightarrow \dfrac{1 + \sqrt{65}}{4} \text{ or } \dfrac{1 - \sqrt{65}}{4}$

Hence, roots of the given equation are $\dfrac{1 + \sqrt{65}}{4} , \dfrac{1 - \sqrt{65}}{4}$.

(iii) $5x^2 - 6\sqrt{5}x + 9 = 0$

The given equation is $5x^2 - 6\sqrt{5}x + 9 = 0$

Comparing it with ax² + bx + c = 0
a= 5, b = -6√5, c = 9

$\therefore \text{Discriminant } = b^2 - 4ac \\[1em] = (-6\sqrt{5})^2 - 4 \times 5 \times 9 \\[1em] = 180 - 180 \\[1em] = 0$

Since, Discriminant = 0, hence equation has two equal and real roots.

By using the formula , x = $\dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$\Rightarrow \dfrac{-(-6\sqrt{5}) ± \sqrt{(-6\sqrt{5})^2 - 4 \times 5 \times 9}}{2 \times 5} \\[1em] \Rightarrow \dfrac{6\sqrt{5} ± \sqrt{180 - 180}}{10} \\[1em] \Rightarrow \dfrac{6\sqrt{5} + \sqrt{0}}{10} \text{ or } \dfrac{6\sqrt{5} - \sqrt{0}}{10} \\[1em] \Rightarrow \dfrac{6\sqrt{5}}{10} \text{ or } \dfrac{6\sqrt{5}}{10} \\[1em] \Rightarrow \dfrac{3\sqrt{5}}{5} \text{ or } \dfrac{3\sqrt{5}}{5} \\[1em] \dfrac{3}{\sqrt{5}} \text{ or } \dfrac{3}{\sqrt{5}}$

Hence, roots of the given equation are $\dfrac{3}{\sqrt{5}}, \dfrac{3}{\sqrt{5}}$.

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

The given equation is $\sqrt{3}x^2 - 2x - \sqrt{3} = 0$.

Comparing it with ax² + bx + c = 0
a= $\sqrt{3}$, b = -2, c = -$\sqrt{3}$

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

Since, Discriminant > 0 , hence equation has two distinct and real roots.

By using the formula , x = $\dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$\Rightarrow \dfrac{-(-2) ± \sqrt{(-2)^2 - 4 \times \sqrt{3} \times -\sqrt{3}}}{2 \times \sqrt{3}} \\[1em] \Rightarrow \dfrac{2 ± \sqrt{4 + 12}}{2\sqrt{3}} \\[1em] \Rightarrow \dfrac{2 + \sqrt{16}}{2\sqrt{3}} \text{ or } \dfrac{2 - \sqrt{16}}{2\sqrt{3}} \\[1em] \Rightarrow \dfrac{2 + 4}{2\sqrt{3}} \text{ or } \dfrac{2- 4}{2\sqrt{3}} \\[1em] \Rightarrow \dfrac{6}{2\sqrt{3}} \text{ or } -\dfrac{2}{2\sqrt{3}} \\[1em] \sqrt{3} \text{ or } -\dfrac{1}{\sqrt{3}}$

Hence, roots of the given equation are $\sqrt{3}, -\dfrac{1}{\sqrt{3}}$.