Solve 4x^2-13x+7leq0

$$4 { x }^{ 2 } -13x+7 \leq 0$$

Solve for x

$$x\in \begin{bmatrix}\frac{13-\sqrt{57}}{8},\frac{\sqrt{57}+13}{8}\end{bmatrix}$$

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.

$$4x^{2}-13x+7=0$$

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, -13 for b, and 7 for c in the quadratic formula.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 4\times 7}}{2\times 4}$$

Do the calculations.

$$x=\frac{13±\sqrt{57}}{8}$$

Solve the equation x=\frac{13±\sqrt{57}}{8} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{57}+13}{8} x=\frac{13-\sqrt{57}}{8}$$

Rewrite the inequality by using the obtained solutions.

$$4\left(x-\frac{\sqrt{57}+13}{8}\right)\left(x-\frac{13-\sqrt{57}}{8}\right)\leq 0$$

For the product to be ≤0, one of the values x-\frac{\sqrt{57}+13}{8} and x-\frac{13-\sqrt{57}}{8} has to be ≥0 and the other has to be ≤0. Consider the case when x-\frac{\sqrt{57}+13}{8}\geq 0 and x-\frac{13-\sqrt{57}}{8}\leq 0.

$$x-\frac{\sqrt{57}+13}{8}\geq 0 x-\frac{13-\sqrt{57}}{8}\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\frac{\sqrt{57}+13}{8}\leq 0 and x-\frac{13-\sqrt{57}}{8}\geq 0.

$$x-\frac{13-\sqrt{57}}{8}\geq 0 x-\frac{\sqrt{57}+13}{8}\leq 0$$

The solution satisfying both inequalities is x\in \left[\frac{13-\sqrt{57}}{8},\frac{\sqrt{57}+13}{8}\right].

$$x\in \begin{bmatrix}\frac{13-\sqrt{57}}{8},\frac{\sqrt{57}+13}{8}\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}\frac{13-\sqrt{57}}{8},\frac{\sqrt{57}+13}{8}\end{bmatrix}$$