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

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

Solve for x

$$x=4$$

Solution Steps

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

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

Square both sides of the equation.

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

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

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

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

$$x=16-8\sqrt{x}+\left(\sqrt{x}\right)^{2}$$

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

$$x=16-8\sqrt{x}+x$$

Add 8\sqrt{x} to both sides.

$$x+8\sqrt{x}=16+x$$

Subtract x from both sides.

$$x+8\sqrt{x}-x=16$$

Combine x and -x to get 0.

$$8\sqrt{x}=16$$

Divide both sides by 8.

$$\sqrt{x}=\frac{16}{8}$$

Divide 16 by 8 to get 2.

$$\sqrt{x}=2$$

Square both sides of the equation.

$$x=4$$

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

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

Simplify. The value x=4 satisfies the equation.

$$4=4$$

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

$$x=4$$