Solve x+sqrt{x+6}=0

$$x + \sqrt { x + 6 } = 0$$

Solve for x

$$x=-2$$

Solution Steps

Subtract x from both sides of the equation.

$$\sqrt{x+6}=-x$$

Square both sides of the equation.

$$\left(\sqrt{x+6}\right)^{2}=\left(-x\right)^{2}$$

Calculate \sqrt{x+6} to the power of 2 and get x+6.

$$x+6=\left(-x\right)^{2}$$

Expand \left(-x\right)^{2}.

$$x+6=\left(-1\right)^{2}x^{2}$$

Calculate -1 to the power of 2 and get 1.

$$x+6=1x^{2}$$

Reorder the terms.

$$x+6=x^{2}$$

Subtract x^{2} from both sides.

$$x+6-x^{2}=0$$

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

$$-x^{2}+x+6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=1 ab=-6=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 1.

$$a=3 b=-2$$

Rewrite -x^{2}+x+6 as \left(-x^{2}+3x\right)+\left(-2x+6\right).

$$\left(-x^{2}+3x\right)+\left(-2x+6\right)$$

Factor out -x in the first and -2 in the second group.

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

Factor out common term x-3 by using distributive property.

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

To find equation solutions, solve x-3=0 and -x-2=0.

$$x=3 x=-2$$

Substitute 3 for x in the equation x+\sqrt{x+6}=0.

$$3+\sqrt{3+6}=0$$

Simplify. The value x=3 does not satisfy the equation.

$$6=0$$

Substitute -2 for x in the equation x+\sqrt{x+6}=0.

$$-2+\sqrt{-2+6}=0$$

Simplify. The value x=-2 satisfies the equation.

$$0=0$$

Equation \sqrt{x+6}=-x has a unique solution.

$$x=-2$$