Solve 10(1000-x)(1+0.2%x)geq12x

$$10(1000-x)(1+0.2 \% x) \geq 12x$$

Solve for x

$$x\in \begin{bmatrix}-50\sqrt{201}-50,50\sqrt{201}-50\end{bmatrix}$$

Steps Using the Quadratic Formula

Expand \frac{0.2}{100} by multiplying both numerator and the denominator by 10.

$$10\left(1000-x\right)\left(1+\frac{2}{1000}x\right)\geq 12x$$

Reduce the fraction \frac{2}{1000} to lowest terms by extracting and canceling out 2.

$$10\left(1000-x\right)\left(1+\frac{1}{500}x\right)\geq 12x$$

Use the distributive property to multiply 10 by 1000-x.

$$\left(10000-10x\right)\left(1+\frac{1}{500}x\right)\geq 12x$$

Use the distributive property to multiply 10000-10x by 1+\frac{1}{500}x and combine like terms.

$$10000+10x-\frac{1}{50}x^{2}\geq 12x$$

Subtract 12x from both sides.

$$10000+10x-\frac{1}{50}x^{2}-12x\geq 0$$

Combine 10x and -12x to get -2x.

$$10000-2x-\frac{1}{50}x^{2}\geq 0$$

Multiply the inequality by -1 to make the coefficient of the highest power in 10000-2x-\frac{1}{50}x^{2} positive. Since -1 is negative, the inequality direction is changed.

$$-10000+2x+\frac{1}{50}x^{2}\leq 0$$

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.

$$-10000+2x+\frac{1}{50}x^{2}=0$$

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 \frac{1}{50} for a, 2 for b, and -10000 for c in the quadratic formula.

$$x=\frac{-2±\sqrt{2^{2}-4\times \frac{1}{50}\left(-10000\right)}}{\frac{1}{50}\times 2}$$

Do the calculations.

$$x=\frac{-2±2\sqrt{201}}{\frac{1}{25}}$$

Solve the equation x=\frac{-2±2\sqrt{201}}{\frac{1}{25}} when ± is plus and when ± is minus.

$$x=50\sqrt{201}-50 x=-50\sqrt{201}-50$$

Rewrite the inequality by using the obtained solutions.

$$\frac{1}{50}\left(x-\left(50\sqrt{201}-50\right)\right)\left(x-\left(-50\sqrt{201}-50\right)\right)\leq 0$$

For the product to be ≤0, one of the values x-\left(50\sqrt{201}-50\right) and x-\left(-50\sqrt{201}-50\right) has to be ≥0 and the other has to be ≤0. Consider the case when x-\left(50\sqrt{201}-50\right)\geq 0 and x-\left(-50\sqrt{201}-50\right)\leq 0.

$$x-\left(50\sqrt{201}-50\right)\geq 0 x-\left(-50\sqrt{201}-50\right)\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\left(50\sqrt{201}-50\right)\leq 0 and x-\left(-50\sqrt{201}-50\right)\geq 0.

$$x-\left(-50\sqrt{201}-50\right)\geq 0 x-\left(50\sqrt{201}-50\right)\leq 0$$

The solution satisfying both inequalities is x\in \left[-50\sqrt{201}-50,50\sqrt{201}-50\right].

$$x\in \begin{bmatrix}-50\sqrt{201}-50,50\sqrt{201}-50\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-50\sqrt{201}-50,50\sqrt{201}-50\end{bmatrix}$$