Solve sqrt{left(x-2right)^2}+3x=6

$$\sqrt{ { \left(x-2 \right) }^{ 2 } } +3x = 6$$

Solve for x

$$x=2$$

Solution Steps

Subtract 3x from both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=6-3x$$

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

$$\sqrt{x^{2}-4x+4}=6-3x$$

Square both sides of the equation.

$$\left(\sqrt{x^{2}-4x+4}\right)^{2}=\left(6-3x\right)^{2}$$

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

$$x^{2}-4x+4=\left(6-3x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6-3x\right)^{2}.

$$x^{2}-4x+4=36-36x+9x^{2}$$

Subtract 36 from both sides.

$$x^{2}-4x+4-36=-36x+9x^{2}$$

Subtract 36 from 4 to get -32.

$$x^{2}-4x-32=-36x+9x^{2}$$

Add 36x to both sides.

$$x^{2}-4x-32+36x=9x^{2}$$

Combine -4x and 36x to get 32x.

$$x^{2}+32x-32=9x^{2}$$

Subtract 9x^{2} from both sides.

$$x^{2}+32x-32-9x^{2}=0$$

Combine x^{2} and -9x^{2} to get -8x^{2}.

$$-8x^{2}+32x-32=0$$

Divide both sides by 8.

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

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.

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

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 4.

$$a=2 b=2$$

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

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

Factor out -x in the first and 2 in the second group.

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

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

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

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

$$x=2 x=2$$

Substitute 2 for x in the equation \sqrt{\left(x-2\right)^{2}}+3x=6.

$$\sqrt{\left(2-2\right)^{2}}+3\times 2=6$$

Simplify. The value x=2 satisfies the equation.

$$6=6$$

Substitute 2 for x in the equation \sqrt{\left(x-2\right)^{2}}+3x=6.

$$\sqrt{\left(2-2\right)^{2}}+3\times 2=6$$

Simplify. The value x=2 satisfies the equation.

$$6=6$$

List all solutions of \sqrt{x^{2}-4x+4}=6-3x.

$$x=2 x=2$$