Solve 2x^2-13x+11leq0

$$2 x ^ { 2 } - 13 x + 11 \leq 0$$

Solve for x

$$x\in \begin{bmatrix}1,\frac{11}{2}\end{bmatrix}$$

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.

$$2x^{2}-13x+11=0$$

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 2 for a, -13 for b, and 11 for c in the quadratic formula.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 2\times 11}}{2\times 2}$$

Do the calculations.

$$x=\frac{13±9}{4}$$

Solve the equation x=\frac{13±9}{4} when ± is plus and when ± is minus.

$$x=\frac{11}{2} x=1$$

Rewrite the inequality by using the obtained solutions.

$$2\left(x-\frac{11}{2}\right)\left(x-1\right)\leq 0$$

For the product to be ≤0, one of the values x-\frac{11}{2} and x-1 has to be ≥0 and the other has to be ≤0. Consider the case when x-\frac{11}{2}\geq 0 and x-1\leq 0.

$$x-\frac{11}{2}\geq 0 x-1\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\frac{11}{2}\leq 0 and x-1\geq 0.

$$x-1\geq 0 x-\frac{11}{2}\leq 0$$

The solution satisfying both inequalities is x\in \left[1,\frac{11}{2}\right].

$$x\in \begin{bmatrix}1,\frac{11}{2}\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}1,\frac{11}{2}\end{bmatrix}$$