Solve x^2-7x+4>0

$$x ^ { 2 } - 7 x + 4 > 0$$

Solve for x

$$x\in \left(-\infty,\frac{7-\sqrt{33}}{2}\right)\cup \left(\frac{\sqrt{33}+7}{2},\infty\right)$$

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.

$$x^{2}-7x+4=0$$

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 1 for a, -7 for b, and 4 for c in the quadratic formula.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 1\times 4}}{2}$$

Do the calculations.

$$x=\frac{7±\sqrt{33}}{2}$$

Solve the equation x=\frac{7±\sqrt{33}}{2} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{33}+7}{2} x=\frac{7-\sqrt{33}}{2}$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-\frac{\sqrt{33}+7}{2}\right)\left(x-\frac{7-\sqrt{33}}{2}\right)>0$$

For the product to be positive, x-\frac{\sqrt{33}+7}{2} and x-\frac{7-\sqrt{33}}{2} have to be both negative or both positive. Consider the case when x-\frac{\sqrt{33}+7}{2} and x-\frac{7-\sqrt{33}}{2} are both negative.

$$x-\frac{\sqrt{33}+7}{2}<0 x-\frac{7-\sqrt{33}}{2}<0$$

The solution satisfying both inequalities is x<\frac{7-\sqrt{33}}{2}.

$$x<\frac{7-\sqrt{33}}{2}$$

Consider the case when x-\frac{\sqrt{33}+7}{2} and x-\frac{7-\sqrt{33}}{2} are both positive.

$$x-\frac{7-\sqrt{33}}{2}>0 x-\frac{\sqrt{33}+7}{2}>0$$

The solution satisfying both inequalities is x>\frac{\sqrt{33}+7}{2}.

$$x>\frac{\sqrt{33}+7}{2}$$

The final solution is the union of the obtained solutions.

$$x<\frac{7-\sqrt{33}}{2}\text{; }x>\frac{\sqrt{33}+7}{2}$$