Solve sqrt{2x+3}=sqrt{x+1}+1

$$\sqrt{ 2x+3 } = \sqrt{ x+1 } +1$$

Solve for x

$$x=3$$

$$x=-1$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{2x+3}\right)^{2}=\left(\sqrt{x+1}+1\right)^{2}$$

Calculate \sqrt{2x+3} to the power of 2 and get 2x+3.

$$2x+3=\left(\sqrt{x+1}+1\right)^{2}$$

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

$$2x+3=\left(\sqrt{x+1}\right)^{2}+2\sqrt{x+1}+1$$

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

$$2x+3=x+1+2\sqrt{x+1}+1$$

Add 1 and 1 to get 2.

$$2x+3=x+2+2\sqrt{x+1}$$

Subtract x+2 from both sides of the equation.

$$2x+3-\left(x+2\right)=2\sqrt{x+1}$$

To find the opposite of x+2, find the opposite of each term.

$$2x+3-x-2=2\sqrt{x+1}$$

Combine 2x and -x to get x.

$$x+3-2=2\sqrt{x+1}$$

Subtract 2 from 3 to get 1.

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

Square both sides of the equation.

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

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

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

Expand \left(2\sqrt{x+1}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Use the distributive property to multiply 4 by x+1.

$$x^{2}+2x+1=4x+4$$

Subtract 4x from both sides.

$$x^{2}+2x+1-4x=4$$

Combine 2x and -4x to get -2x.

$$x^{2}-2x+1=4$$

Subtract 4 from both sides.

$$x^{2}-2x+1-4=0$$

Subtract 4 from 1 to get -3.

$$x^{2}-2x-3=0$$

To solve the equation, factor x^{2}-2x-3 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-2 ab=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-3 b=1$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-3\right)\left(x+1\right)$$

To find equation solutions, solve x-3=0 and x+1=0.

$$x=3 x=-1$$

Substitute 3 for x in the equation \sqrt{2x+3}=\sqrt{x+1}+1.

$$\sqrt{2\times 3+3}=\sqrt{3+1}+1$$

Simplify. The value x=3 satisfies the equation.

$$3=3$$

Substitute -1 for x in the equation \sqrt{2x+3}=\sqrt{x+1}+1.

$$\sqrt{2\left(-1\right)+3}=\sqrt{-1+1}+1$$

Simplify. The value x=-1 satisfies the equation.

$$1=1$$

List all solutions of \sqrt{2x+3}=\sqrt{x+1}+1.

$$x=3 x=-1$$