Given,
5 x + 4 3 = 2 3 x 2 \dfrac{5}{x} + 4\sqrt{3} = \dfrac{2\sqrt{3}}{x^2} x 5 + 4 3 = x 2 2 3
Substituting 1 x \dfrac{1}{x} x 1
⇒ 5 t + 4 3 = 2 3 t 2 ⇒ 2 3 t 2 − 5 t − 4 3 = 0 ⇒ 2 3 t 2 − 8 t + 3 t − 4 3 = 0 ⇒ 2 t ( 3 t − 4 ) + 3 ( 3 t − 4 ) = 0 ⇒ ( 2 t + 3 ) ( 3 t − 4 ) = 0 ⇒ 2 t + 3 = 0 or 3 t − 4 = 0 ⇒ 2 t = − 3 or 3 t = 4 ⇒ t = − 3 2 or t = 4 3 ⇒ 1 x = − 3 2 or 1 x = 4 3 ⇒ x = − 2 3 or x = 3 4 . \Rightarrow 5t + 4\sqrt{3} = 2\sqrt{3}t^2 \\[1em] \Rightarrow 2\sqrt{3}t^2 - 5t - 4\sqrt{3} = 0 \\[1em] \Rightarrow 2\sqrt{3}t^2 - 8t + 3t - 4\sqrt{3}= 0 \\[1em] \Rightarrow 2t(\sqrt{3}t - 4) + \sqrt{3}(\sqrt{3}t - 4) = 0 \\[1em] \Rightarrow (2t + \sqrt{3})(\sqrt{3}t - 4) = 0 \\[1em] \Rightarrow 2t + \sqrt{3} = 0 \text{ or } \sqrt{3}t - 4 = 0 \\[1em] \Rightarrow 2t = -\sqrt{3} \text{ or } \sqrt{3}t = 4 \\[1em] \Rightarrow t = -\dfrac{\sqrt{3}}{2} \text{ or } t = \dfrac{4}{\sqrt{3}} \\[1em] \Rightarrow \dfrac{1}{x} = -\dfrac{\sqrt{3}}{2} \text{ or } \dfrac{1}{x} = \dfrac{4}{\sqrt{3}} \\[1em] \Rightarrow x = -\dfrac{2}{\sqrt{3}} \text{ or } x = \dfrac{\sqrt{3}}{4}. ⇒ 5 t + 4 3 = 2 3 t 2 ⇒ 2 3 t 2 − 5 t − 4 3 = 0 ⇒ 2 3 t 2 − 8 t + 3 t − 4 3 = 0 ⇒ 2 t ( 3 t − 4 ) + 3 ( 3 t − 4 ) = 0 ⇒ ( 2 t + 3 ) ( 3 t − 4 ) = 0 ⇒ 2 t + 3 = 0 or 3 t − 4 = 0 ⇒ 2 t = − 3 or 3 t = 4 ⇒ t = − 2 3 or t = 3 4 ⇒ x 1 = − 2 3 or x 1 = 3 4 ⇒ x = − 3 2 or x = 4 3 .
Hence, x = − 2 3 or x = 3 4 . -\dfrac{2}{\sqrt{3}} \text{ or } x = \dfrac{\sqrt{3}}{4}. − 3 2 or x = 4 3 .
