Solve sqrt{x+4}-sqrt{3x}=-2

$$\sqrt{ x+4 } - \sqrt{ 3x } = -2$$

Solve for x

$$x=12$$

Solution Steps

Subtract -\sqrt{3x} from both sides of the equation.

$$\sqrt{x+4}=-2+\sqrt{3x}$$

Square both sides of the equation.

$$\left(\sqrt{x+4}\right)^{2}=\left(-2+\sqrt{3x}\right)^{2}$$

Calculate \sqrt{x+4} to the power of 2 and get x+4.

$$x+4=\left(-2+\sqrt{3x}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-2+\sqrt{3x}\right)^{2}.

$$x+4=4-4\sqrt{3x}+\left(\sqrt{3x}\right)^{2}$$

Calculate \sqrt{3x} to the power of 2 and get 3x.

$$x+4=4-4\sqrt{3x}+3x$$

Subtract 4+3x from both sides of the equation.

$$x+4-\left(4+3x\right)=-4\sqrt{3x}$$

To find the opposite of 4+3x, find the opposite of each term.

$$x+4-4-3x=-4\sqrt{3x}$$

Subtract 4 from 4 to get 0.

$$x-3x=-4\sqrt{3x}$$

Combine x and -3x to get -2x.

$$-2x=-4\sqrt{3x}$$

Square both sides of the equation.

$$\left(-2x\right)^{2}=\left(-4\sqrt{3x}\right)^{2}$$

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

$$\left(-2\right)^{2}x^{2}=\left(-4\sqrt{3x}\right)^{2}$$

Calculate -2 to the power of 2 and get 4.

$$4x^{2}=\left(-4\sqrt{3x}\right)^{2}$$

Expand \left(-4\sqrt{3x}\right)^{2}.

$$4x^{2}=\left(-4\right)^{2}\left(\sqrt{3x}\right)^{2}$$

Calculate -4 to the power of 2 and get 16.

$$4x^{2}=16\left(\sqrt{3x}\right)^{2}$$

Calculate \sqrt{3x} to the power of 2 and get 3x.

$$4x^{2}=16\times 3x$$

Multiply 16 and 3 to get 48.

$$4x^{2}=48x$$

Subtract 48x from both sides.

$$4x^{2}-48x=0$$

Factor out x.

$$x\left(4x-48\right)=0$$

To find equation solutions, solve x=0 and 4x-48=0.

$$x=0 x=12$$

Substitute 0 for x in the equation \sqrt{x+4}-\sqrt{3x}=-2.

$$\sqrt{0+4}-\sqrt{3\times 0}=-2$$

Simplify. The value x=0 does not satisfy the equation because the left and the right hand side have opposite signs.

$$2=-2$$

Substitute 12 for x in the equation \sqrt{x+4}-\sqrt{3x}=-2.

$$\sqrt{12+4}-\sqrt{3\times 12}=-2$$

Simplify. The value x=12 satisfies the equation.

$$-2=-2$$

Equation \sqrt{x+4}=\sqrt{3x}-2 has a unique solution.

$$x=12$$