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

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

Solve for x

$$x=5$$

Solution Steps

Square both sides of the equation.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{x+4}+\sqrt{x-4}\right)^{2}.

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

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

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

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

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

Combine x and x to get 2x.

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

Subtract 4 from 4 to get 0.

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

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

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

Subtract 2x from both sides of the equation.

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

Combine 4x and -2x to get 2x.

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

Square both sides of the equation.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

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

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

Use the distributive property to multiply 4 by x+4.

$$\left(4x+16\right)\left(x-4\right)=\left(2x-4\right)^{2}$$

Apply the distributive property by multiplying each term of 4x+16 by each term of x-4.

$$4x^{2}-16x+16x-64=\left(2x-4\right)^{2}$$

Combine -16x and 16x to get 0.

$$4x^{2}-64=\left(2x-4\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-4\right)^{2}.

$$4x^{2}-64=4x^{2}-16x+16$$

Subtract 4x^{2} from both sides.

$$4x^{2}-64-4x^{2}=-16x+16$$

Combine 4x^{2} and -4x^{2} to get 0.

$$-64=-16x+16$$

Swap sides so that all variable terms are on the left hand side.

$$-16x+16=-64$$

Subtract 16 from both sides.

$$-16x=-64-16$$

Subtract 16 from -64 to get -80.

$$-16x=-80$$

Divide both sides by -16.

$$x=\frac{-80}{-16}$$

Divide -80 by -16 to get 5.

$$x=5$$

Substitute 5 for x in the equation \sqrt{x+4}+\sqrt{x-4}=\sqrt{4x-4}.

$$\sqrt{5+4}+\sqrt{5-4}=\sqrt{4\times 5-4}$$

Simplify. The value x=5 satisfies the equation.

$$4=4$$

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

$$x=5$$