Solve 3x^2-1=x-1

$$3 x ^ { 2 } - 1 = x - 1$$

Solve for x

$$x=\frac{1}{3}\approx 0.333333333$$

$$x=0$$

Steps Using Factoring

Subtract x from both sides.

$$3x^{2}-1-x=-1$$

Add 1 to both sides.

$$3x^{2}-1-x+1=0$$

Add -1 and 1 to get 0.

$$3x^{2}-x=0$$

Factor out x.

$$x\left(3x-1\right)=0$$

To find equation solutions, solve x=0 and 3x-1=0.

$$x=0 x=\frac{1}{3}$$

Steps Using the Quadratic Formula

Subtract x from both sides.

$$3x^{2}-1-x=-1$$

Add 1 to both sides.

$$3x^{2}-1-x+1=0$$

Add -1 and 1 to get 0.

$$3x^{2}-x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -1 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1}}{2\times 3}$$

Take the square root of 1.

$$x=\frac{-\left(-1\right)±1}{2\times 3}$$

The opposite of -1 is 1.

$$x=\frac{1±1}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{1±1}{6}$$

Now solve the equation x=\frac{1±1}{6} when ± is plus. Add 1 to 1.

$$x=\frac{2}{6}$$

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

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

Now solve the equation x=\frac{1±1}{6} when ± is minus. Subtract 1 from 1.

$$x=\frac{0}{6}$$

Divide 0 by 6.

$$x=0$$

The equation is now solved.

$$x=\frac{1}{3} x=0$$

Steps for Completing the Square

Subtract x from both sides.

$$3x^{2}-1-x=-1$$

Add 1 to both sides.

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

Add -1 and 1 to get 0.

$$3x^{2}-x=0$$

Divide both sides by 3.

$$\frac{3x^{2}-x}{3}=\frac{0}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{1}{3}x=\frac{0}{3}$$

Divide 0 by 3.

$$x^{2}-\frac{1}{3}x=0$$

Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=\left(-\frac{1}{6}\right)^{2}$$

Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{1}{36}$$

Factor x^{2}-\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{1}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Simplify.

$$x-\frac{1}{6}=\frac{1}{6} x-\frac{1}{6}=-\frac{1}{6}$$

Add \frac{1}{6} to both sides of the equation.

$$x=\frac{1}{3} x=0$$