Solve x-sqrt{32-4x}=0

$$x - \sqrt { 32 - 4 x } = 0$$

Solve for x

$$x=4$$

Solution Steps

Subtract x from both sides of the equation.

$$-\sqrt{32-4x}=-x$$

Cancel out -1 on both sides.

$$\sqrt{32-4x}=x$$

Square both sides of the equation.

$$\left(\sqrt{32-4x}\right)^{2}=x^{2}$$

Calculate \sqrt{32-4x} to the power of 2 and get 32-4x.

$$32-4x=x^{2}$$

Subtract x^{2} from both sides.

$$32-4x-x^{2}=0$$

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

$$-x^{2}-4x+32=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+32. To find a and b, set up a system to be solved.

$$a+b=-4 ab=-32=-32$$

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 -32.

$$1,-32 2,-16 4,-8$$

Calculate the sum for each pair.

$$1-32=-31 2-16=-14 4-8=-4$$

The solution is the pair that gives sum -4.

$$a=4 b=-8$$

Rewrite -x^{2}-4x+32 as \left(-x^{2}+4x\right)+\left(-8x+32\right).

$$\left(-x^{2}+4x\right)+\left(-8x+32\right)$$

Factor out x in the first and 8 in the second group.

$$x\left(-x+4\right)+8\left(-x+4\right)$$

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

$$\left(-x+4\right)\left(x+8\right)$$

To find equation solutions, solve -x+4=0 and x+8=0.

$$x=4 x=-8$$

Substitute 4 for x in the equation x-\sqrt{32-4x}=0.

$$4-\sqrt{32-4\times 4}=0$$

Simplify. The value x=4 satisfies the equation.

$$0=0$$

Substitute -8 for x in the equation x-\sqrt{32-4x}=0.

$$-8-\sqrt{32-4\left(-8\right)}=0$$

Simplify. The value x=-8 does not satisfy the equation.

$$-16=0$$

Equation \sqrt{32-4x}=x has a unique solution.

$$x=4$$