Solve 4x^2-5x+1>0

$$4 { x }^{ 2 } -5x+1 > 0$$

Solve for x

$$x\in \left(-\infty,\frac{1}{4}\right)\cup \left(1,\infty\right)$$

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.

$$4x^{2}-5x+1=0$$

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 4 for a, -5 for b, and 1 for c in the quadratic formula.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 4\times 1}}{2\times 4}$$

Do the calculations.

$$x=\frac{5±3}{8}$$

Solve the equation x=\frac{5±3}{8} when ± is plus and when ± is minus.

$$x=1 x=\frac{1}{4}$$

Rewrite the inequality by using the obtained solutions.

$$4\left(x-1\right)\left(x-\frac{1}{4}\right)>0$$

For the product to be positive, x-1 and x-\frac{1}{4} have to be both negative or both positive. Consider the case when x-1 and x-\frac{1}{4} are both negative.

$$x-1<0 x-\frac{1}{4}<0$$

The solution satisfying both inequalities is x<\frac{1}{4}.

$$x<\frac{1}{4}$$

Consider the case when x-1 and x-\frac{1}{4} are both positive.

$$x-\frac{1}{4}>0 x-1>0$$

The solution satisfying both inequalities is x>1.

$$x>1$$

The final solution is the union of the obtained solutions.

$$x<\frac{1}{4}\text{; }x>1$$