Solve 1+sqrt{y}=sqrt{y+13}

$$1 + \sqrt { y } = \sqrt { y + 13 }$$

Solve for y

$$y=36$$

Solution Steps

Square both sides of the equation.

$$\left(1+\sqrt{y}\right)^{2}=\left(\sqrt{y+13}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+\sqrt{y}\right)^{2}.

$$1+2\sqrt{y}+\left(\sqrt{y}\right)^{2}=\left(\sqrt{y+13}\right)^{2}$$

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

$$1+2\sqrt{y}+y=\left(\sqrt{y+13}\right)^{2}$$

Calculate \sqrt{y+13} to the power of 2 and get y+13.

$$1+2\sqrt{y}+y=y+13$$

Subtract y from both sides.

$$1+2\sqrt{y}+y-y=13$$

Combine y and -y to get 0.

$$1+2\sqrt{y}=13$$

Subtract 1 from both sides.

$$2\sqrt{y}=13-1$$

Subtract 1 from 13 to get 12.

$$2\sqrt{y}=12$$

Divide both sides by 2.

$$\sqrt{y}=\frac{12}{2}$$

Divide 12 by 2 to get 6.

$$\sqrt{y}=6$$

Square both sides of the equation.

$$y=36$$

Substitute 36 for y in the equation 1+\sqrt{y}=\sqrt{y+13}.

$$1+\sqrt{36}=\sqrt{36+13}$$

Simplify. The value y=36 satisfies the equation.

$$7=7$$

Equation \sqrt{y}+1=\sqrt{y+13} has a unique solution.

$$y=36$$