Solve frac{-2sqrt{x-4}}{-2}=x-4/-2

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

Solve for x

$$x=4$$

Solution Steps

Multiply both sides of the equation by -2.

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

Subtract x from both sides.

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

Subtract -x from both sides of the equation.

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

Square both sides of the equation.

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

Expand \left(-2\sqrt{x-4}\right)^{2}.

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

Calculate -2 to the power of 2 and get 4.

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

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

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

Use the distributive property to multiply 4 by x-4.

$$4x-16=\left(-4+x\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-4+x\right)^{2}.

$$4x-16=16-8x+x^{2}$$

Add 8x to both sides.

$$4x-16+8x=16+x^{2}$$

Combine 4x and 8x to get 12x.

$$12x-16=16+x^{2}$$

Subtract x^{2} from both sides.

$$12x-16-x^{2}=16$$

Subtract 16 from both sides.

$$12x-16-x^{2}-16=0$$

Subtract 16 from -16 to get -32.

$$12x-32-x^{2}=0$$

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

$$-x^{2}+12x-32=0$$

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.

$$a+b=12 ab=-\left(-32\right)=32$$

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 32.

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

Calculate the sum for each pair.

$$1+32=33 2+16=18 4+8=12$$

The solution is the pair that gives sum 12.

$$a=8 b=4$$

Rewrite -x^{2}+12x-32 as \left(-x^{2}+8x\right)+\left(4x-32\right).

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

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

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

Factor out common term x-8 by using distributive property.

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

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

$$x=8 x=4$$

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

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

Simplify. The value x=8 does not satisfy the equation because the left and the right hand side have opposite signs.

$$2=-2$$

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

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

Simplify. The value x=4 satisfies the equation.

$$0=0$$

Equation -2\sqrt{x-4}=x-4 has a unique solution.

$$x=4$$