The value of x is
and ![x=\frac{-1-\sqrt{37}}{12}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-1-%5Csqrt%7B37%7D%7D%7B12%7D)
Step-by-step explanation:
The equation is ![\frac{1}{x}-\frac{2}{3}=4 x](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%7D-%5Cfrac%7B2%7D%7B3%7D%3D4%20x)
Subtracting by
on both sides,
![\frac{1}{x}-\frac{2}{3}-4 x=0](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%7D-%5Cfrac%7B2%7D%7B3%7D-4%20x%3D0)
Taking LCM,
![\frac{3-12 x^{2}-2 x}{3 x}=0](https://tex.z-dn.net/?f=%5Cfrac%7B3-12%20x%5E%7B2%7D-2%20x%7D%7B3%20x%7D%3D0)
Multiplying by 3x on both sides,
![-12 x^{2}-2 x+3=0](https://tex.z-dn.net/?f=-12%20x%5E%7B2%7D-2%20x%2B3%3D0)
Dividing by (-) on both sides,
![12 x^{2}+2 x-3=0](https://tex.z-dn.net/?f=12%20x%5E%7B2%7D%2B2%20x-3%3D0)
Using quadratic formula, we can solve for x.
![\begin{aligned}x &=\frac{-2 \pm \sqrt{2^{2}-4 \cdot 12 \cdot(-3)}}{2 \cdot 12} \\&=\frac{-2 \pm \sqrt{4+144}}{2 \cdot 144} \\&=\frac{-2 \pm \sqrt{148}}{24} \\&=\frac{-2 \pm 2 \sqrt{37}}{24}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Dx%20%26%3D%5Cfrac%7B-2%20%5Cpm%20%5Csqrt%7B2%5E%7B2%7D-4%20%5Ccdot%2012%20%5Ccdot%28-3%29%7D%7D%7B2%20%5Ccdot%2012%7D%20%5C%5C%26%3D%5Cfrac%7B-2%20%5Cpm%20%5Csqrt%7B4%2B144%7D%7D%7B2%20%5Ccdot%20144%7D%20%5C%5C%26%3D%5Cfrac%7B-2%20%5Cpm%20%5Csqrt%7B148%7D%7D%7B24%7D%20%5C%5C%26%3D%5Cfrac%7B-2%20%5Cpm%202%20%5Csqrt%7B37%7D%7D%7B24%7D%5Cend%7Baligned%7D)
Taking out common term 2, we get,
![\begin{array}{l}{x=\frac{-2(1 \pm \sqrt{37})}{24}} \\{x=\frac{-1 \pm \sqrt{37}}{12}}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7Bx%3D%5Cfrac%7B-2%281%20%5Cpm%20%5Csqrt%7B37%7D%29%7D%7B24%7D%7D%20%5C%5C%7Bx%3D%5Cfrac%7B-1%20%5Cpm%20%5Csqrt%7B37%7D%7D%7B12%7D%7D%5Cend%7Barray%7D)
Thus, the value of x is
and ![x=\frac{-1-\sqrt{37}}{12}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-1-%5Csqrt%7B37%7D%7D%7B12%7D)