Solve sqrt{8-x}=x-2

$$\sqrt{ 8-x } =x-2$$

Solve for x

$$x=4$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{8-x}\right)^{2}=\left(x-2\right)^{2}$$

Calculate \sqrt{8-x} to the power of 2 and get 8-x.

$$8-x=\left(x-2\right)^{2}$$

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

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

Subtract x^{2} from both sides.

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

Add 4x to both sides.

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

Combine -x and 4x to get 3x.

$$8+3x-x^{2}=4$$

Subtract 4 from both sides.

$$8+3x-x^{2}-4=0$$

Subtract 4 from 8 to get 4.

$$4+3x-x^{2}=0$$

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

$$-x^{2}+3x+4=0$$

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.

$$a+b=3 ab=-4=-4$$

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. List all such integer pairs that give product -4.

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

Calculate the sum for each pair.

$$-1+4=3 -2+2=0$$

The solution is the pair that gives sum 3.

$$a=4 b=-1$$

Rewrite -x^{2}+3x+4 as \left(-x^{2}+4x\right)+\left(-x+4\right).

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

Factor out -x in the first and -1 in the second group.

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

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{8-x}=x-2.

$$\sqrt{8-4}=4-2$$

Simplify. The value x=4 satisfies the equation.

$$2=2$$

Substitute -1 for x in the equation \sqrt{8-x}=x-2.

$$\sqrt{8-\left(-1\right)}=-1-2$$

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

$$3=-3$$

Equation \sqrt{8-x}=x-2 has a unique solution.

$$x=4$$