Solve sqrt{x}+1=sqrt{x+5}

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

Solve for x

$$x=4$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x}+1\right)^{2}=\left(\sqrt{x+5}\right)^{2}$$

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

$$\left(\sqrt{x}\right)^{2}+2\sqrt{x}+1=\left(\sqrt{x+5}\right)^{2}$$

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

$$x+2\sqrt{x}+1=\left(\sqrt{x+5}\right)^{2}$$

Calculate \sqrt{x+5} to the power of 2 and get x+5.

$$x+2\sqrt{x}+1=x+5$$

Subtract x from both sides.

$$x+2\sqrt{x}+1-x=5$$

Combine x and -x to get 0.

$$2\sqrt{x}+1=5$$

Subtract 1 from both sides.

$$2\sqrt{x}=5-1$$

Subtract 1 from 5 to get 4.

$$2\sqrt{x}=4$$

Divide both sides by 2.

$$\sqrt{x}=\frac{4}{2}$$

Divide 4 by 2 to get 2.

$$\sqrt{x}=2$$

Square both sides of the equation.

$$x=4$$

Substitute 4 for x in the equation \sqrt{x}+1=\sqrt{x+5}.

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

Simplify. The value x=4 satisfies the equation.

$$3=3$$

Equation \sqrt{x}+1=\sqrt{x+5} has a unique solution.

$$x=4$$