Find the quadratic equation whose solution set is:

(i) {2, -3}

(ii) $\Big{-3, \dfrac{2}{5}\Big}$

(iii) $\Big{\dfrac{2}{5}, -\dfrac{1}{2}\Big}$

(i) Since, {2, -3} is solution set.

It means 2 and -3 are roots of the equation,

∴ x = 2 or x = -3

⇒ x - 2 = 0 or x + 3 = 0

⇒ (x - 2)(x + 3) = 0

⇒ (x² + 3x - 2x - 6) = 0

⇒ x² + x - 6 = 0.

Hence, quadratic equation with solution set {2, -3} is x² + x - 6 = 0.

(ii) Since, $\Big{-3, \dfrac{2}{5}\Big}$ is solution set.

It means -3 and $\dfrac{2}{5}$ are roots of the equation,

∴ x = -3 or x = $\dfrac{2}{5}$

⇒ x = -3 or 5x = 2

⇒ x + 3 = 0 or 5x - 2 = 0

⇒ (x + 3)(5x - 2) = 0

⇒ (5x² - 2x + 15x - 6) = 0

⇒ 5x² + 13x - 6 = 0.

Hence, quadratic equation with solution set $\Big{-3, \dfrac{2}{5}\Big}$ is 5x² + 13x - 6 = 0.

(iii) Since, $\Big{\dfrac{2}{5}, -\dfrac{1}{2}\Big}$ is solution set.

It means $\dfrac{2}{5}$ and $-\dfrac{1}{2}$ are roots of the equation,

∴ x = $\dfrac{2}{5}$ or x = $-\dfrac{1}{2}$

⇒ 5x = 2 or 2x = -1

⇒ 5x - 2 = 0 or 2x + 1 = 0

⇒ (5x - 2)(2x + 1) = 0

⇒ (10x² + 5x - 4x - 2) = 0

⇒ 10x² + x - 2 = 0.

Hence, quadratic equation with solution set $\Big{\dfrac{2}{5}, -\dfrac{1}{2}\Big}$ is 10x² + x - 2 = 0.