Solve 0.02x^2+100x+80000leq0

$$0.02 x ^ { 2 } + 100 x + 80000 \leq 0$$

Solve for x

$$x\in \begin{bmatrix}-4000,-1000\end{bmatrix}$$

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.

$$0.02x^{2}+100x+80000=0$$

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 0.02 for a, 100 for b, and 80000 for c in the quadratic formula.

$$x=\frac{-100±\sqrt{100^{2}-4\times 0.02\times 80000}}{0.02\times 2}$$

Do the calculations.

$$x=\frac{-100±60}{0.04}$$

Solve the equation x=\frac{-100±60}{0.04} when ± is plus and when ± is minus.

$$x=-1000 x=-4000$$

Rewrite the inequality by using the obtained solutions.

$$0.02\left(x+1000\right)\left(x+4000\right)\leq 0$$

For the product to be ≤0, one of the values x+1000 and x+4000 has to be ≥0 and the other has to be ≤0. Consider the case when x+1000\geq 0 and x+4000\leq 0.

$$x+1000\geq 0 x+4000\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+1000\leq 0 and x+4000\geq 0.

$$x+4000\geq 0 x+1000\leq 0$$

The solution satisfying both inequalities is x\in \left[-4000,-1000\right].

$$x\in \begin{bmatrix}-4000,-1000\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-4000,-1000\end{bmatrix}$$