Question

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Answer 1

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$\frac{ {x}^{3} + 3x }{ {3x}^{2} + 1} = \frac{14}{13} \\ \\ 13( {x}^{3} + 3x) = 14( {3x}^{2} + 1) \\ \\ 13 {x}^{3} + 39x = 42 {x}^{2} + 14 \\ \\ 13 {x}^{3} + 39x - 42 {x}^{2} - 14 = 0$

Factorise it and solve.

Sorry i don't have enough battery to do it

Answer 2

$\mathrm{Given\ that\ \ \dfrac{x^3+3x}{3x^2+1}=\dfrac{14}{13}} \\ \\ \\ \implies \mathrm{ 13\big(x^3+3x\big)=14\big(3x^2+1\big)} \\ \\ \implies \mathrm{13x^3 + 39x = 42x^2+14} \\ \\ \implies \mathrm{13x^3-42x^2+39x-14=0} \\ \\ \implies \mathrm{13x^3 - 26x^2-16x^2+32x+7x-14=0} \\ \\ \implies \mathrm{13x^2(x-2)-16x(x-2)+7(x-2)=0}\\ \\ \implies \mathrm{Taking \ (x-2)\ common\ we\ get\ :} \\ \\ \implies \mathrm{(x-2)(13x^2-16x+7)=0} \\ \\ \mathrm{So, x=2 \ is\ one\ real\ solution\ for \ the\ above\ equation.}$

$\mathrm{Now\ solving\ the\ quadratic\ we\ get\ other\ two\ solutions\ of\ x.} \\ \\ \mathrm{Using\ quadratic\ equation\ formula\ on\ 13x^2-16x+7=0 \ we \ get:}\\ \\ \mathrm{x = \dfrac{-(-16) \pm \sqrt{(-16)^2 -4(13)(7)}}{2(13)}} \\ \\ \\ \implies \mathrm{x= \dfrac{8}{13}\ \pm \ \dfrac{3\sqrt{3}\ i}{13} } \\ \\ \\ \boxed{\mathrm{Therefore\ the\ values\ of\ x\ are:\ \ 2\ , \dfrac{8}{13}\ \pm \ \dfrac{3\sqrt{3}\ i}{13} }}$

Hope this helps : )