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

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

Solve for x

$$x=0$$

Solution Steps

Subtract -2x from both sides of the equation.

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

Combine 2x and 2x to get 4x.

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

Square both sides of the equation.

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

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

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

Calculate 4 to the power of 2 and get 16.

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

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

$$16x^{2}=x^{2}-3x$$

Subtract x^{2} from both sides.

$$16x^{2}-x^{2}=-3x$$

Combine 16x^{2} and -x^{2} to get 15x^{2}.

$$15x^{2}=-3x$$

Add 3x to both sides.

$$15x^{2}+3x=0$$

Factor out x.

$$x\left(15x+3\right)=0$$

To find equation solutions, solve x=0 and 15x+3=0.

$$x=0 x=-\frac{1}{5}$$

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

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

Simplify. The value x=0 satisfies the equation.

$$0=0$$

Substitute -\frac{1}{5} for x in the equation 2x=-2x+\sqrt{x^{2}-3x}.

$$2\left(-\frac{1}{5}\right)=-2\left(-\frac{1}{5}\right)+\sqrt{\left(-\frac{1}{5}\right)^{2}-3\left(-\frac{1}{5}\right)}$$

Simplify. The value x=-\frac{1}{5} does not satisfy the equation because the left and the right hand side have opposite signs.

$$-\frac{2}{5}=\frac{6}{5}$$

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

$$x=0$$