Solve sqrt{3x+1}+1=3x

$$\sqrt { 3 x + 1 } + 1 = 3 x$$

Solve for x

$$x=1$$

Solution Steps

Subtract 1 from both sides of the equation.

$$\sqrt{3x+1}=3x-1$$

Square both sides of the equation.

$$\left(\sqrt{3x+1}\right)^{2}=\left(3x-1\right)^{2}$$

Calculate \sqrt{3x+1} to the power of 2 and get 3x+1.

$$3x+1=\left(3x-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-1\right)^{2}.

$$3x+1=9x^{2}-6x+1$$

Subtract 9x^{2} from both sides.

$$3x+1-9x^{2}=-6x+1$$

Add 6x to both sides.

$$3x+1-9x^{2}+6x=1$$

Combine 3x and 6x to get 9x.

$$9x+1-9x^{2}=1$$

Subtract 1 from both sides.

$$9x+1-9x^{2}-1=0$$

Subtract 1 from 1 to get 0.

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

Factor out x.

$$x\left(9-9x\right)=0$$

To find equation solutions, solve x=0 and 9-9x=0.

$$x=0 x=1$$

Substitute 0 for x in the equation \sqrt{3x+1}+1=3x.

$$\sqrt{3\times 0+1}+1=3\times 0$$

Simplify. The value x=0 does not satisfy the equation.

$$2=0$$

Substitute 1 for x in the equation \sqrt{3x+1}+1=3x.

$$\sqrt{3\times 1+1}+1=3\times 1$$

Simplify. The value x=1 satisfies the equation.

$$3=3$$

Equation \sqrt{3x+1}=3x-1 has a unique solution.

$$x=1$$