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

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

Solve for x

$$x=0$$

Solution Steps

Subtract -3x from both sides of the equation.

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

Square both sides of the equation.

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

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

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

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

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

Subtract 1 from both sides.

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

Subtract 1 from 1 to get 0.

$$x=6x+9x^{2}$$

Subtract 6x from both sides.

$$x-6x=9x^{2}$$

Combine x and -6x to get -5x.

$$-5x=9x^{2}$$

Subtract 9x^{2} from both sides.

$$-5x-9x^{2}=0$$

Factor out x.

$$x\left(-5-9x\right)=0$$

To find equation solutions, solve x=0 and -5-9x=0.

$$x=0 x=-\frac{5}{9}$$

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

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

Simplify. The value x=0 satisfies the equation.

$$1=1$$

Substitute -\frac{5}{9} for x in the equation \sqrt{x+1}-3x=1.

$$\sqrt{-\frac{5}{9}+1}-3\left(-\frac{5}{9}\right)=1$$

Simplify. The value x=-\frac{5}{9} does not satisfy the equation.

$$\frac{7}{3}=1$$

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

$$x=0$$