Solve sqrt{x}+x+1=0

$$\sqrt{ x } +x+1=0$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{3}i-1}{2}\approx -0.5-0.866025404i$$

Solution Steps

Subtract x+1 from both sides of the equation.

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

To find the opposite of x+1, find the opposite of each term.

$$\sqrt{x}=-x-1$$

Square both sides of the equation.

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

Calculate \sqrt{x} to the power of 2 and get x.

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

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

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

Subtract x^{2} from both sides.

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

Subtract 2x from both sides.

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

Combine x and -2x to get -x.

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

Subtract 1 from both sides.

$$-x-x^{2}-1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-x^{2}-x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -1 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Multiply -4 times -1.

$$x=\frac{-\left(-1\right)±\sqrt{1+4\left(-1\right)}}{2\left(-1\right)}$$

Multiply 4 times -1.

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

Add 1 to -4.

$$x=\frac{-\left(-1\right)±\sqrt{-3}}{2\left(-1\right)}$$

Take the square root of -3.

$$x=\frac{-\left(-1\right)±\sqrt{3}i}{2\left(-1\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{3}i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{1±\sqrt{3}i}{-2}$$

Now solve the equation x=\frac{1±\sqrt{3}i}{-2} when ± is plus. Add 1 to i\sqrt{3}.

$$x=\frac{1+\sqrt{3}i}{-2}$$

Divide 1+i\sqrt{3} by -2.

$$x=\frac{-\sqrt{3}i-1}{2}$$

Now solve the equation x=\frac{1±\sqrt{3}i}{-2} when ± is minus. Subtract i\sqrt{3} from 1.

$$x=\frac{-\sqrt{3}i+1}{-2}$$

Divide 1-i\sqrt{3} by -2.

$$x=\frac{-1+\sqrt{3}i}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{3}i-1}{2} x=\frac{-1+\sqrt{3}i}{2}$$

Substitute \frac{-\sqrt{3}i-1}{2} for x in the equation \sqrt{x}+x+1=0.

$$\sqrt{\frac{-\sqrt{3}i-1}{2}}+\frac{-\sqrt{3}i-1}{2}+1=0$$

Simplify. The value x=\frac{-\sqrt{3}i-1}{2} satisfies the equation.

$$0=0$$

Substitute \frac{-1+\sqrt{3}i}{2} for x in the equation \sqrt{x}+x+1=0.

$$\sqrt{\frac{-1+\sqrt{3}i}{2}}+\frac{-1+\sqrt{3}i}{2}+1=0$$

Simplify. The value x=\frac{-1+\sqrt{3}i}{2} does not satisfy the equation.

$$1+i\times 3^{\frac{1}{2}}=0$$

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

$$x=\frac{-\sqrt{3}i-1}{2}$$