Solve {r}{3x^2-4x}{+1>0}

$$\left. \begin{array} { r } { 3 x ^ { 2 } - 4 x } \\ { + 1 > 0 } \end{array} \right.$$

Solve for x

$$x\in \left(-\infty,\frac{1}{3}\right)\cup \left(1,\infty\right)$$

Steps Using the Quadratic Formula

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$3x^{2}-4x+1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 3 for a, -4 for b, and 1 for c in the quadratic formula.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 3\times 1}}{2\times 3}$$

Do the calculations.

$$x=\frac{4±2}{6}$$

Solve the equation x=\frac{4±2}{6} when ± is plus and when ± is minus.

$$x=1 x=\frac{1}{3}$$

Rewrite the inequality by using the obtained solutions.

$$3\left(x-1\right)\left(x-\frac{1}{3}\right)>0$$

For the product to be positive, x-1 and x-\frac{1}{3} have to be both negative or both positive. Consider the case when x-1 and x-\frac{1}{3} are both negative.

$$x-1<0 x-\frac{1}{3}<0$$

The solution satisfying both inequalities is x<\frac{1}{3}.

$$x<\frac{1}{3}$$

Consider the case when x-1 and x-\frac{1}{3} are both positive.

$$x-\frac{1}{3}>0 x-1>0$$

The solution satisfying both inequalities is x>1.

$$x>1$$

The final solution is the union of the obtained solutions.

$$x<\frac{1}{3}\text{; }x>1$$