Solve sqrt{8-2x}=x

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

Solve for x

$$x=2$$

Solution Steps

Square both sides of the equation.

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

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

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

Subtract x^{2} from both sides.

$$8-2x-x^{2}=0$$

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+8. To find a and b, set up a system to be solved.

$$a+b=-2 ab=-8=-8$$

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. List all such integer pairs that give product -8.

$$1,-8 2,-4$$

Calculate the sum for each pair.

$$1-8=-7 2-4=-2$$

The solution is the pair that gives sum -2.

$$a=2 b=-4$$

Rewrite -x^{2}-2x+8 as \left(-x^{2}+2x\right)+\left(-4x+8\right).

$$\left(-x^{2}+2x\right)+\left(-4x+8\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(-x+2\right)+4\left(-x+2\right)$$

Factor out common term -x+2 by using distributive property.

$$\left(-x+2\right)\left(x+4\right)$$

To find equation solutions, solve -x+2=0 and x+4=0.

$$x=2 x=-4$$

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

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

Simplify. The value x=2 satisfies the equation.

$$2=2$$

Substitute -4 for x in the equation \sqrt{8-2x}=x.

$$\sqrt{8-2\left(-4\right)}=-4$$

Simplify. The value x=-4 does not satisfy the equation because the left and the right hand side have opposite signs.

$$4=-4$$

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

$$x=2$$