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

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

Solve for x

$$x=7$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

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

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

Add 2 and 1 to get 3.

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

Subtract 2x+3 from both sides of the equation.

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

To find the opposite of 2x+3, find the opposite of each term.

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

Combine x and -2x to get -x.

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

Subtract 3 from 2 to get -1.

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

Square both sides of the equation.

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

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

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

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

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

Calculate -2 to the power of 2 and get 4.

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

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

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

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

$$x^{2}+2x+1=8x+8$$

Subtract 8x from both sides.

$$x^{2}+2x+1-8x=8$$

Combine 2x and -8x to get -6x.

$$x^{2}-6x+1=8$$

Subtract 8 from both sides.

$$x^{2}-6x+1-8=0$$

Subtract 8 from 1 to get -7.

$$x^{2}-6x-7=0$$

To solve the equation, factor x^{2}-6x-7 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.

$$a+b=-6 ab=-7$$

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.

$$a=-7 b=1$$

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

$$\left(x-7\right)\left(x+1\right)$$

To find equation solutions, solve x-7=0 and x+1=0.

$$x=7 x=-1$$

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

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

Simplify. The value x=7 satisfies the equation.

$$3=3$$

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

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

Simplify. The value x=-1 does not satisfy the equation because the left and the right hand side have opposite signs.

$$1=-1$$

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

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

Simplify. The value x=7 satisfies the equation.

$$3=3$$

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

$$x=7$$