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Mathematics

Solve for x, if 5x+43=23x2\dfrac{5}{x} + 4\sqrt{3} = \dfrac{2\sqrt{3}}{x^2}, x ≠ 0

Quadratic Equations

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Answer

Given,

5x+43=23x2\dfrac{5}{x} + 4\sqrt{3} = \dfrac{2\sqrt{3}}{x^2}

Substituting 1x\dfrac{1}{x} = t in the given equation, we get :

⇒5t+43=23t2⇒23t2−5t−43=0⇒23t2−8t+3t−43=0⇒2t(3t−4)+3(3t−4)=0⇒(2t+3)(3t−4)=0⇒2t+3=0 or 3t−4=0⇒2t=−3 or 3t=4⇒t=−32 or t=43⇒1x=−32 or 1x=43⇒x=−23 or x=34.\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}.

Hence, x = −23 or x=34.-\dfrac{2}{\sqrt{3}} \text{ or } x = \dfrac{\sqrt{3}}{4}.

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