Solve sqrt{x}+sqrt{x+1}=2

$$\sqrt { x } + \sqrt { x + 1 } = 2$$

Solve for x

$$x=\frac{9}{16}=0.5625$$

Solution Steps

Subtract \sqrt{x+1} from both sides of the equation.

$$\sqrt{x}=2-\sqrt{x+1}$$

Square both sides of the equation.

$$\left(\sqrt{x}\right)^{2}=\left(2-\sqrt{x+1}\right)^{2}$$

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

$$x=\left(2-\sqrt{x+1}\right)^{2}$$

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

$$x=4-4\sqrt{x+1}+\left(\sqrt{x+1}\right)^{2}$$

Calculate \sqrt{x+1} to the power of 2 and get x+1.

$$x=4-4\sqrt{x+1}+x+1$$

Add 4 and 1 to get 5.

$$x=5-4\sqrt{x+1}+x$$

Add 4\sqrt{x+1} to both sides.

$$x+4\sqrt{x+1}=5+x$$

Subtract x from both sides.

$$x+4\sqrt{x+1}-x=5$$

Combine x and -x to get 0.

$$4\sqrt{x+1}=5$$

Divide both sides by 4.

$$\sqrt{x+1}=\frac{5}{4}$$

Square both sides of the equation.

$$x+1=\frac{25}{16}$$

Subtract 1 from both sides of the equation.

$$x+1-1=\frac{25}{16}-1$$

Subtracting 1 from itself leaves 0.

$$x=\frac{25}{16}-1$$

Subtract 1 from \frac{25}{16}.

$$x=\frac{9}{16}$$

Substitute \frac{9}{16} for x in the equation \sqrt{x}+\sqrt{x+1}=2.

$$\sqrt{\frac{9}{16}}+\sqrt{\frac{9}{16}+1}=2$$

Simplify. The value x=\frac{9}{16} satisfies the equation.

$$2=2$$

Equation \sqrt{x}=-\sqrt{x+1}+2 has a unique solution.

$$x=\frac{9}{16}$$