Solve sqrt{2x+7}=sqrt{x+3}+1

$$\sqrt { 2 x + 7 } = \sqrt { x + 3 } + 1$$

Solve for x

$$x=1$$

$$x=-3$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{2x+7}\right)^{2}=\left(\sqrt{x+3}+1\right)^{2}$$

Calculate \sqrt{2x+7} to the power of 2 and get 2x+7.

$$2x+7=\left(\sqrt{x+3}+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{x+3}+1\right)^{2}.

$$2x+7=\left(\sqrt{x+3}\right)^{2}+2\sqrt{x+3}+1$$

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

$$2x+7=x+3+2\sqrt{x+3}+1$$

Add 3 and 1 to get 4.

$$2x+7=x+4+2\sqrt{x+3}$$

Subtract x+4 from both sides of the equation.

$$2x+7-\left(x+4\right)=2\sqrt{x+3}$$

To find the opposite of x+4, find the opposite of each term.

$$2x+7-x-4=2\sqrt{x+3}$$

Combine 2x and -x to get x.

$$x+7-4=2\sqrt{x+3}$$

Subtract 4 from 7 to get 3.

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

Square both sides of the equation.

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

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

$$x^{2}+6x+9=\left(2\sqrt{x+3}\right)^{2}$$

Expand \left(2\sqrt{x+3}\right)^{2}.

$$x^{2}+6x+9=2^{2}\left(\sqrt{x+3}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$x^{2}+6x+9=4\left(\sqrt{x+3}\right)^{2}$$

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

$$x^{2}+6x+9=4\left(x+3\right)$$

Use the distributive property to multiply 4 by x+3.

$$x^{2}+6x+9=4x+12$$

Subtract 4x from both sides.

$$x^{2}+6x+9-4x=12$$

Combine 6x and -4x to get 2x.

$$x^{2}+2x+9=12$$

Subtract 12 from both sides.

$$x^{2}+2x+9-12=0$$

Subtract 12 from 9 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 positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=3$$

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

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

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

$$x=1 x=-3$$

Substitute 1 for x in the equation \sqrt{2x+7}=\sqrt{x+3}+1.

$$\sqrt{2\times 1+7}=\sqrt{1+3}+1$$

Simplify. The value x=1 satisfies the equation.

$$3=3$$

Substitute -3 for x in the equation \sqrt{2x+7}=\sqrt{x+3}+1.

$$\sqrt{2\left(-3\right)+7}=\sqrt{-3+3}+1$$

Simplify. The value x=-3 satisfies the equation.

$$1=1$$

List all solutions of \sqrt{2x+7}=\sqrt{x+3}+1.

$$x=1 x=-3$$