Solve sqrt{y+3}=sqrt{y}+sqrt{3}

$$\sqrt { y + 3 } = \sqrt { y } + \sqrt { 3 }$$

Solve for y

$$y=0$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{y+3}\right)^{2}=\left(\sqrt{y}+\sqrt{3}\right)^{2}$$

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

$$y+3=\left(\sqrt{y}+\sqrt{3}\right)^{2}$$

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

$$y+3=\left(\sqrt{y}\right)^{2}+2\sqrt{y}\sqrt{3}+\left(\sqrt{3}\right)^{2}$$

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

$$y+3=y+2\sqrt{y}\sqrt{3}+\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$y+3=y+2\sqrt{y}\sqrt{3}+3$$

Subtract y from both sides.

$$y+3-y=2\sqrt{y}\sqrt{3}+3$$

Combine y and -y to get 0.

$$3=2\sqrt{y}\sqrt{3}+3$$

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

$$2\sqrt{y}\sqrt{3}+3=3$$

Subtract 3 from both sides.

$$2\sqrt{y}\sqrt{3}=3-3$$

Subtract 3 from 3 to get 0.

$$2\sqrt{y}\sqrt{3}=0$$

Divide both sides by 2\sqrt{3}.

$$\frac{2\sqrt{3}\sqrt{y}}{2\sqrt{3}}=\frac{0}{2\sqrt{3}}$$

Dividing by 2\sqrt{3} undoes the multiplication by 2\sqrt{3}.

$$\sqrt{y}=\frac{0}{2\sqrt{3}}$$

Divide 0 by 2\sqrt{3}.

$$\sqrt{y}=0$$

Square both sides of the equation.

$$y=0$$

Substitute 0 for y in the equation \sqrt{y+3}=\sqrt{y}+\sqrt{3}.

$$\sqrt{0+3}=\sqrt{0}+\sqrt{3}$$

Simplify. The value y=0 satisfies the equation.

$$3^{\frac{1}{2}}=3^{\frac{1}{2}}$$

Equation \sqrt{y+3}=\sqrt{y}+\sqrt{3} has a unique solution.

$$y=0$$