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

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

Solve for x

$$x=0$$

Solution Steps

Square both sides of the equation.

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

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

$$9x^{2}+12x+4=\left(\sqrt{\left(3x+2\right)^{2}+x^{2}}\right)^{2}$$

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

$$9x^{2}+12x+4=\left(\sqrt{9x^{2}+12x+4+x^{2}}\right)^{2}$$

Combine 9x^{2} and x^{2} to get 10x^{2}.

$$9x^{2}+12x+4=\left(\sqrt{10x^{2}+12x+4}\right)^{2}$$

Calculate \sqrt{10x^{2}+12x+4} to the power of 2 and get 10x^{2}+12x+4.

$$9x^{2}+12x+4=10x^{2}+12x+4$$

Subtract 10x^{2} from both sides.

$$9x^{2}+12x+4-10x^{2}=12x+4$$

Combine 9x^{2} and -10x^{2} to get -x^{2}.

$$-x^{2}+12x+4=12x+4$$

Subtract 12x from both sides.

$$-x^{2}+12x+4-12x=4$$

Combine 12x and -12x to get 0.

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

Subtract 4 from both sides.

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

Subtract 4 from 4 to get 0.

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

Divide both sides by -1. Zero divided by any non-zero number gives zero.

$$x^{2}=0$$

Take the square root of both sides of the equation.

$$x=0 x=0$$

The equation is now solved. Solutions are the same.

$$x=0$$

Substitute 0 for x in the equation 3x+2=\sqrt{\left(3x+2\right)^{2}+x^{2}}.

$$3\times 0+2=\sqrt{\left(3\times 0+2\right)^{2}+0^{2}}$$

Simplify. The value x=0 satisfies the equation.

$$2=2$$

Equation 3x+2=\sqrt{\left(3x+2\right)^{2}+x^{2}} has a unique solution.

$$x=0$$