Solve (x+1)^2-2(x+1)-1leq0

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

Solve for x

$$x\in \begin{bmatrix}-\sqrt{2},\sqrt{2}\end{bmatrix}$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$x^{2}+2x+1-2\left(x+1\right)-1\leq 0$$

Use the distributive property to multiply -2 by x+1.

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

Combine 2x and -2x to get 0.

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

Subtract 2 from 1 to get -1.

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

Subtract 1 from -1 to get -2.

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

Add 2 to both sides.

$$x^{2}\leq 2$$

Calculate the square root of 2 and get \sqrt{2}. Rewrite 2 as \left(\sqrt{2}\right)^{2}.

$$x^{2}\leq \left(\sqrt{2}\right)^{2}$$

Inequality holds for |x|\leq \sqrt{2}.

$$|x|\leq \sqrt{2}$$

Rewrite |x|\leq \sqrt{2} as x\in \left[-\sqrt{2},\sqrt{2}\right].

$$x\in \begin{bmatrix}-\sqrt{2},\sqrt{2}\end{bmatrix}$$