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

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

Solve for x

$$x=0$$

Solution Steps

Square both sides of the equation.

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

To raise a power to another power, multiply the exponents. Multiply 1 and 2 to get 2.

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

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

$$9x^{2}+6x+1=\left(\sqrt{x^{2}+1}\right)^{2}$$

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

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

Subtract x^{2} from both sides.

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

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

$$8x^{2}+6x+1=1$$

Subtract 1 from both sides.

$$8x^{2}+6x+1-1=0$$

Subtract 1 from 1 to get 0.

$$8x^{2}+6x=0$$

Factor out x.

$$x\left(8x+6\right)=0$$

To find equation solutions, solve x=0 and 8x+6=0.

$$x=0 x=-\frac{3}{4}$$

Substitute 0 for x in the equation \left(3x+1\right)^{1}=\sqrt{x^{2}+1}.

$$\left(3\times 0+1\right)^{1}=\sqrt{0^{2}+1}$$

Simplify. The value x=0 satisfies the equation.

$$1=1$$

Substitute -\frac{3}{4} for x in the equation \left(3x+1\right)^{1}=\sqrt{x^{2}+1}.

$$\left(3\left(-\frac{3}{4}\right)+1\right)^{1}=\sqrt{\left(-\frac{3}{4}\right)^{2}+1}$$

Simplify. The value x=-\frac{3}{4} does not satisfy the equation because the left and the right hand side have opposite signs.

$$-\frac{5}{4}=\frac{5}{4}$$

Equation \left(3x+1\right)^{1}=\sqrt{x^{2}+1} has a unique solution.

$$x=0$$