Solve 10x^2-33x+20leq0

$$10 x ^ { 2 } - 33 x + 20 \leq 0$$

Solve for x

$$x\in \begin{bmatrix}\frac{4}{5},\frac{5}{2}\end{bmatrix}$$

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.

$$10x^{2}-33x+20=0$$

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 10 for a, -33 for b, and 20 for c in the quadratic formula.

$$x=\frac{-\left(-33\right)±\sqrt{\left(-33\right)^{2}-4\times 10\times 20}}{2\times 10}$$

Do the calculations.

$$x=\frac{33±17}{20}$$

Solve the equation x=\frac{33±17}{20} when ± is plus and when ± is minus.

$$x=\frac{5}{2} x=\frac{4}{5}$$

Rewrite the inequality by using the obtained solutions.

$$10\left(x-\frac{5}{2}\right)\left(x-\frac{4}{5}\right)\leq 0$$

For the product to be ≤0, one of the values x-\frac{5}{2} and x-\frac{4}{5} has to be ≥0 and the other has to be ≤0. Consider the case when x-\frac{5}{2}\geq 0 and x-\frac{4}{5}\leq 0.

$$x-\frac{5}{2}\geq 0 x-\frac{4}{5}\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\frac{5}{2}\leq 0 and x-\frac{4}{5}\geq 0.

$$x-\frac{4}{5}\geq 0 x-\frac{5}{2}\leq 0$$

The solution satisfying both inequalities is x\in \left[\frac{4}{5},\frac{5}{2}\right].

$$x\in \begin{bmatrix}\frac{4}{5},\frac{5}{2}\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}\frac{4}{5},\frac{5}{2}\end{bmatrix}$$