How do you solve the quadratic $x-4/(3x)=-1/3$ using any method?

Question

2713 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-02T15:20:45

$x=-4/3, 1$

Given that

$x-4/{3x}=-1/3$

$x-4/{3x}+1/3=0$

$\frac{3x^2-4+x}{3x}=0$

$\frac{3x^2+x-4}{3x}=0$

$\frac{3x^2+4x-3x-4}{3x}=0$

$\frac{x(3x+4)-(3x+4)}{3x}=0$

$\frac{(3x+4)(x-1)}{3x}=0$

$(3x+4)(x-1)=0\quad (\forall \ x\ne 0)$

$x=-4/3, 1$

How do you solve the quadratic x-4/(3x)=-1/3x-4/(3x)=-1/3 using any method?