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

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

Solve for x

$$x=7$$

Solution Steps

Subtract 1 from both sides of the equation.

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

Square both sides of the equation.

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

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

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

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

$$x^{2}-2x+1=4x+8$$

Subtract 4x from both sides.

$$x^{2}-2x+1-4x=8$$

Combine -2x and -4x 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 x=2\sqrt{x+2}+1.

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

Simplify. The value x=7 satisfies the equation.

$$7=7$$

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

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

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

$$-1=3$$

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

$$x=7$$