Solve sqrt{-x}=x+2

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

Solve for x

$$x=-1$$

Solution Steps

Square both sides of the equation.

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

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

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

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

$$-x=x^{2}+4x+4$$

Subtract x^{2} from both sides.

$$-x-x^{2}=4x+4$$

Subtract 4x from both sides.

$$-x-x^{2}-4x=4$$

Subtract 4 from both sides.

$$-x-x^{2}-4x-4=0$$

Combine -x and -4x to get -5x.

$$-5x-x^{2}-4=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}-5x-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=-5 ab=-\left(-4\right)=4$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -5.

$$a=-1 b=-4$$

Rewrite -x^{2}-5x-4 as \left(-x^{2}-x\right)+\left(-4x-4\right).

$$\left(-x^{2}-x\right)+\left(-4x-4\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(-x-1\right)+4\left(-x-1\right)$$

Factor out common term -x-1 by using distributive property.

$$\left(-x-1\right)\left(x+4\right)$$

To find equation solutions, solve -x-1=0 and x+4=0.

$$x=-1 x=-4$$

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

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

Simplify. The value x=-1 satisfies the equation.

$$1=1$$

Substitute -4 for x in the equation \sqrt{-x}=x+2.

$$\sqrt{-\left(-4\right)}=-4+2$$

Simplify. The value x=-4 does not satisfy the equation because the left and the right hand side have opposite signs.

$$2=-2$$

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

$$x=-1$$