Solve sqrt{2x+5}=sqrt{x+2}+1

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

Solve for x

$$x=2$$

$$x=-2$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

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

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

Add 2 and 1 to get 3.

$$2x+5=x+3+2\sqrt{x+2}$$

Subtract x+3 from both sides of the equation.

$$2x+5-\left(x+3\right)=2\sqrt{x+2}$$

To find the opposite of x+3, find the opposite of each term.

$$2x+5-x-3=2\sqrt{x+2}$$

Combine 2x and -x to get x.

$$x+5-3=2\sqrt{x+2}$$

Subtract 3 from 5 to get 2.

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

Square both sides of the equation.

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

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

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

Expand \left(2\sqrt{x+2}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Use the distributive property to multiply 4 by x+2.

$$x^{2}+4x+4=4x+8$$

Subtract 4x from both sides.

$$x^{2}+4x+4-4x=8$$

Combine 4x and -4x to get 0.

$$x^{2}+4=8$$

Subtract 8 from both sides.

$$x^{2}+4-8=0$$

Subtract 8 from 4 to get -4.

$$x^{2}-4=0$$

Consider x^{2}-4. Rewrite x^{2}-4 as x^{2}-2^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-2\right)\left(x+2\right)=0$$

To find equation solutions, solve x-2=0 and x+2=0.

$$x=2 x=-2$$

Substitute 2 for x in the equation \sqrt{2x+5}=\sqrt{x+2}+1.

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

Simplify. The value x=2 satisfies the equation.

$$3=3$$

Substitute -2 for x in the equation \sqrt{2x+5}=\sqrt{x+2}+1.

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

Simplify. The value x=-2 satisfies the equation.

$$1=1$$

List all solutions of \sqrt{2x+5}=\sqrt{x+2}+1.

$$x=2 x=-2$$