Solve x=2+sqrt{x}

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

Solve for x

$$x=4$$

Solution Steps

Subtract 2 from both sides of the equation.

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

Square both sides of the equation.

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

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

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

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

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

Subtract x from both sides.

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

Combine -4x and -x to get -5x.

$$x^{2}-5x+4=0$$

To solve the equation, factor x^{2}-5x+4 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.

$$a+b=-5 ab=4$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -5.

$$a=-4 b=-1$$

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

$$\left(x-4\right)\left(x-1\right)$$

To find equation solutions, solve x-4=0 and x-1=0.

$$x=4 x=1$$

Substitute 4 for x in the equation x=2+\sqrt{x}.

$$4=2+\sqrt{4}$$

Simplify. The value x=4 satisfies the equation.

$$4=4$$

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

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

Simplify. The value x=1 does not satisfy the equation.

$$1=3$$

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

$$x=4$$