Solve sqrt{y}+sqrt{y+2}=3

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

Solve for y

$$y = \frac{49}{36} = 1\frac{13}{36} \approx 1.361111111$$

Solution Steps

Subtract \sqrt{y+2} from both sides of the equation.

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

Square both sides of the equation.

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

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

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

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

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

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

$$y=9-6\sqrt{y+2}+y+2$$

Add 9 and 2 to get 11.

$$y=11-6\sqrt{y+2}+y$$

Add 6\sqrt{y+2} to both sides.

$$y+6\sqrt{y+2}=11+y$$

Subtract y from both sides.

$$y+6\sqrt{y+2}-y=11$$

Combine y and -y to get 0.

$$6\sqrt{y+2}=11$$

Divide both sides by 6.

$$\sqrt{y+2}=\frac{11}{6}$$

Square both sides of the equation.

$$y+2=\frac{121}{36}$$

Subtract 2 from both sides of the equation.

$$y+2-2=\frac{121}{36}-2$$

Subtracting 2 from itself leaves 0.

$$y=\frac{121}{36}-2$$

Subtract 2 from \frac{121}{36}.

$$y=\frac{49}{36}$$

Substitute \frac{49}{36} for y in the equation \sqrt{y}+\sqrt{y+2}=3.

$$\sqrt{\frac{49}{36}}+\sqrt{\frac{49}{36}+2}=3$$

Simplify. The value y=\frac{49}{36} satisfies the equation.

$$3=3$$

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

$$y=\frac{49}{36}$$