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

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

Solve for x

$$x=-\frac{3}{4}=-0.75$$

$$x=0$$

Steps Using Factoring

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

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

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.

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

Steps Using the Quadratic Formula

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

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 6 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}}}{2\times 8}$$

Take the square root of 6^{2}.

$$x=\frac{-6±6}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-6±6}{16}$$

Now solve the equation x=\frac{-6±6}{16} when ± is plus. Add -6 to 6.

$$x=\frac{0}{16}$$

Divide 0 by 16.

$$x=0$$

Now solve the equation x=\frac{-6±6}{16} when ± is minus. Subtract 6 from -6.

$$x=-\frac{12}{16}$$

Reduce the fraction \frac{-12}{16} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

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

Steps for Completing the Square

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

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

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.

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

Subtract 1 from 1 to get 0.

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

Divide both sides by 8.

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

Dividing by 8 undoes the multiplication by 8.

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

Reduce the fraction \frac{6}{8} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{3}{4}x=\frac{0}{8}$$

Divide 0 by 8.

$$x^{2}+\frac{3}{4}x=0$$

Divide \frac{3}{4}, the coefficient of the x term, by 2 to get \frac{3}{8}. Then add the square of \frac{3}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{4}x+\left(\frac{3}{8}\right)^{2}=\left(\frac{3}{8}\right)^{2}$$

Square \frac{3}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{4}x+\frac{9}{64}=\frac{9}{64}$$

Factor x^{2}+\frac{3}{4}x+\frac{9}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{8}\right)^{2}=\frac{9}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{8}\right)^{2}}=\sqrt{\frac{9}{64}}$$

Simplify.

$$x+\frac{3}{8}=\frac{3}{8} x+\frac{3}{8}=-\frac{3}{8}$$

Subtract \frac{3}{8} from both sides of the equation.

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