Solve sqrt{x}=x-2

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

Solve for x

$$x=4$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Subtract x^{2} from both sides.

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

Add 4x to both sides.

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

Combine x and 4x to get 5x.

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

Subtract 4 from both sides.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=5 ab=-\left(-4\right)=4$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. 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 -x^{2}+5x-4 as \left(-x^{2}+4x\right)+\left(x-4\right).

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

Factor out -x in -x^{2}+4x.

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

Factor out common term x-4 by using distributive property.

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

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

$$x=4 x=1$$

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

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

Simplify. The value x=4 satisfies the equation.

$$2=2$$

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

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

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

$$1=-1$$

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

$$x=4$$