Solve sqrt{2x^2+2x-1}=-x-1

$$\sqrt { 2 x ^ { 2 } + 2 x - 1 } = - x - 1$$

Solve for x

$$x=-\sqrt{2}\approx -1.414213562$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{2x^{2}+2x-1}\right)^{2}=\left(-x-1\right)^{2}$$

Calculate \sqrt{2x^{2}+2x-1} to the power of 2 and get 2x^{2}+2x-1.

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

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

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

Calculate -x to the power of 2 and get x^{2}.

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

Multiply -2 and -1 to get 2.

$$2x^{2}+2x-1=x^{2}+2x+1$$

Subtract x^{2} from both sides.

$$2x^{2}+2x-1-x^{2}=2x+1$$

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

$$x^{2}+2x-1=2x+1$$

Subtract 2x from both sides.

$$x^{2}+2x-1-2x=1$$

Combine 2x and -2x to get 0.

$$x^{2}-1=1$$

Add 1 to both sides.

$$x^{2}=1+1$$

Add 1 and 1 to get 2.

$$x^{2}=2$$

Take the square root of both sides of the equation.

$$x=\sqrt{2} x=-\sqrt{2}$$

Substitute \sqrt{2} for x in the equation \sqrt{2x^{2}+2x-1}=-x-1.

$$\sqrt{2\left(\sqrt{2}\right)^{2}+2\sqrt{2}-1}=-\sqrt{2}-1$$

Simplify. The value x=\sqrt{2} does not satisfy the equation because the left and the right hand side have opposite signs.

$$2^{\frac{1}{2}}+1=-2^{\frac{1}{2}}-1$$

Substitute -\sqrt{2} for x in the equation \sqrt{2x^{2}+2x-1}=-x-1.

$$\sqrt{2\left(-\sqrt{2}\right)^{2}+2\left(-\sqrt{2}\right)-1}=-\left(-\sqrt{2}\right)-1$$

Simplify. The value x=-\sqrt{2} satisfies the equation.

$$2^{\frac{1}{2}}-1=2^{\frac{1}{2}}-1$$

Equation \sqrt{2x^{2}+2x-1}=-x-1 has a unique solution.

$$x=-\sqrt{2}$$