One root of the quadratic equation 8x² + mx + 15 = 0 is $\dfrac{3}{4}$. Find the value of m. Also, find other root of equation.

Given, $\dfrac{3}{4}$ is root of 8x² + mx + 15 = 0

$\therefore 8\Big(\dfrac{3}{4}\Big)^2 + m \times \dfrac{3}{4} + 15 = 0 \\[1em] \Rightarrow 8 \times \dfrac{9}{16} + \dfrac{3m}{4} + 15 = 0 \\[1em] \Rightarrow \dfrac{9}{2} + \dfrac{3m}{4} + 15 = 0 \\[1em] \Rightarrow \dfrac{3m}{4} + \dfrac{30 + 9}{2} = 0 \\[1em] \Rightarrow \dfrac{3m}{4} = -\dfrac{39}{2} \\[1em] \Rightarrow m = -\dfrac{39}{2} \times \dfrac{4}{3} \\[1em] \Rightarrow m = -26.$

Substituting value of m in equation,

⇒ 8x² + mx + 15 = 0

⇒ 8x² - 26x + 15 = 0

⇒ 8x² - 20x - 6x + 15 = 0

⇒ 4x(2x - 5) - 3(2x - 5) = 0

⇒ (4x - 3)(2x - 5) = 0

⇒ 4x - 3 = 0 or 2x - 5 = 0

⇒ 4x = 3 or 2x = 5

⇒ x = $\dfrac{3}{4}$ or x = $\dfrac{5}{2}$.

Hence, m = -26 and other root = $\dfrac{5}{2}$.