Solve 25x^2-5x-2leq0

$$25 { x }^{ 2 } -5x-2 \leq 0$$

Solve for x

$$x\in \begin{bmatrix}-\frac{1}{5},\frac{2}{5}\end{bmatrix}$$

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.

$$25x^{2}-5x-2=0$$

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 25 for a, -5 for b, and -2 for c in the quadratic formula.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 25\left(-2\right)}}{2\times 25}$$

Do the calculations.

$$x=\frac{5±15}{50}$$

Solve the equation x=\frac{5±15}{50} when ± is plus and when ± is minus.

$$x=\frac{2}{5} x=-\frac{1}{5}$$

Rewrite the inequality by using the obtained solutions.

$$25\left(x-\frac{2}{5}\right)\left(x+\frac{1}{5}\right)\leq 0$$

For the product to be ≤0, one of the values x-\frac{2}{5} and x+\frac{1}{5} has to be ≥0 and the other has to be ≤0. Consider the case when x-\frac{2}{5}\geq 0 and x+\frac{1}{5}\leq 0.

$$x-\frac{2}{5}\geq 0 x+\frac{1}{5}\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\frac{2}{5}\leq 0 and x+\frac{1}{5}\geq 0.

$$x+\frac{1}{5}\geq 0 x-\frac{2}{5}\leq 0$$

The solution satisfying both inequalities is x\in \left[-\frac{1}{5},\frac{2}{5}\right].

$$x\in \begin{bmatrix}-\frac{1}{5},\frac{2}{5}\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-\frac{1}{5},\frac{2}{5}\end{bmatrix}$$