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

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

Solve for x

$$x=\frac{25}{72}\approx 0.347222222$$

Solution Steps

Square both sides of the equation.

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

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

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

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

$$2x+4=9-6\sqrt{2x}+\left(\sqrt{2x}\right)^{2}$$

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

$$2x+4=9-6\sqrt{2x}+2x$$

Add 6\sqrt{2x} to both sides.

$$2x+4+6\sqrt{2x}=9+2x$$

Subtract 2x from both sides.

$$2x+4+6\sqrt{2x}-2x=9$$

Combine 2x and -2x to get 0.

$$4+6\sqrt{2x}=9$$

Subtract 4 from both sides.

$$6\sqrt{2x}=9-4$$

Subtract 4 from 9 to get 5.

$$6\sqrt{2x}=5$$

Divide both sides by 6.

$$\sqrt{2x}=\frac{5}{6}$$

Square both sides of the equation.

$$2x=\frac{25}{36}$$

Divide both sides by 2.

$$\frac{2x}{2}=\frac{\frac{25}{36}}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x=\frac{\frac{25}{36}}{2}$$

Divide \frac{25}{36} by 2.

$$x=\frac{25}{72}$$

Substitute \frac{25}{72} for x in the equation \sqrt{2x+4}=3-\sqrt{2x}.

$$\sqrt{2\times \frac{25}{72}+4}=3-\sqrt{2\times \frac{25}{72}}$$

Simplify. The value x=\frac{25}{72} satisfies the equation.

$$\frac{13}{6}=\frac{13}{6}$$

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

$$x=\frac{25}{72}$$