Solve x^2+2x+1-4x^2+4x-1leq0

$$x ^ { 2 } + 2 x + 1 - 4 x ^ { 2 } + 4 x - 1 \leq 0$$

Solve for x

$$x\in (-\infty,0]\cup [2,\infty)$$

Solution Steps

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+2x+1+4x-1\leq 0$$

Combine 2x and 4x to get 6x.

$$-3x^{2}+6x+1-1\leq 0$$

Subtract 1 from 1 to get 0.

$$-3x^{2}+6x\leq 0$$

Multiply the inequality by -1 to make the coefficient of the highest power in -3x^{2}+6x positive. Since -1 is negative, the inequality direction is changed.

$$3x^{2}-6x\geq 0$$

Factor out x.

$$3x\left(x-2\right)\geq 0$$

For the product to be ≥0, x and x-2 have to be both ≤0 or both ≥0. Consider the case when x and x-2 are both ≤0.

$$x\leq 0 x-2\leq 0$$

The solution satisfying both inequalities is x\leq 0.

$$x\leq 0$$

Consider the case when x and x-2 are both ≥0.

$$x-2\geq 0 x\geq 0$$

The solution satisfying both inequalities is x\geq 2.

$$x\geq 2$$

The final solution is the union of the obtained solutions.

$$x\leq 0\text{; }x\geq 2$$