Solve 3x-2+sqrt{x^2-6x+9}=2(x+3)

$$3 x - 2 + \sqrt { x ^ { 2 } - 6 x + 9 } = 2 ( x + 3 )$$

Solve for x

$$x = \frac{11}{2} = 5\frac{1}{2} = 5.5$$

Solution Steps

Subtract 3x-2 from both sides of the equation.

$$\sqrt{x^{2}-6x+9}=2\left(x+3\right)-\left(3x-2\right)$$

Use the distributive property to multiply 2 by x+3.

$$\sqrt{x^{2}-6x+9}=2x+6-\left(3x-2\right)$$

To find the opposite of 3x-2, find the opposite of each term.

$$\sqrt{x^{2}-6x+9}=2x+6-3x+2$$

Combine 2x and -3x to get -x.

$$\sqrt{x^{2}-6x+9}=-x+6+2$$

Add 6 and 2 to get 8.

$$\sqrt{x^{2}-6x+9}=-x+8$$

Square both sides of the equation.

$$\left(\sqrt{x^{2}-6x+9}\right)^{2}=\left(-x+8\right)^{2}$$

Calculate \sqrt{x^{2}-6x+9} to the power of 2 and get x^{2}-6x+9.

$$x^{2}-6x+9=\left(-x+8\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-x+8\right)^{2}.

$$x^{2}-6x+9=x^{2}-16x+64$$

Subtract x^{2} from both sides.

$$x^{2}-6x+9-x^{2}=-16x+64$$

Combine x^{2} and -x^{2} to get 0.

$$-6x+9=-16x+64$$

Add 16x to both sides.

$$-6x+9+16x=64$$

Combine -6x and 16x to get 10x.

$$10x+9=64$$

Subtract 9 from both sides.

$$10x=64-9$$

Subtract 9 from 64 to get 55.

$$10x=55$$

Divide both sides by 10.

$$x=\frac{55}{10}$$

Reduce the fraction \frac{55}{10} to lowest terms by extracting and canceling out 5.

$$x=\frac{11}{2}$$

Substitute \frac{11}{2} for x in the equation 3x-2+\sqrt{x^{2}-6x+9}=2\left(x+3\right).

$$3\times \frac{11}{2}-2+\sqrt{\left(\frac{11}{2}\right)^{2}-6\times \frac{11}{2}+9}=2\left(\frac{11}{2}+3\right)$$

Simplify. The value x=\frac{11}{2} satisfies the equation.

$$17=17$$

Equation \sqrt{x^{2}-6x+9}=8-x has a unique solution.

$$x=\frac{11}{2}$$