Solve 6x^2=x+1

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

Solve for x

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

$$x=\frac{1}{2}=0.5$$

Steps Using Factoring By Grouping

Subtract x from both sides.

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

Subtract 1 from both sides.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=-1 ab=6\left(-1\right)=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=-3 b=2$$

Rewrite 6x^{2}-x-1 as \left(6x^{2}-3x\right)+\left(2x-1\right).

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

Factor out 3x in 6x^{2}-3x.

$$3x\left(2x-1\right)+2x-1$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(3x+1\right)$$

To find equation solutions, solve 2x-1=0 and 3x+1=0.

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

Steps Using the Quadratic Formula

Subtract x from both sides.

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

Subtract 1 from both sides.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -1 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 6\left(-1\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-1\right)±\sqrt{1-24\left(-1\right)}}{2\times 6}$$

Multiply -24 times -1.

$$x=\frac{-\left(-1\right)±\sqrt{1+24}}{2\times 6}$$

Add 1 to 24.

$$x=\frac{-\left(-1\right)±\sqrt{25}}{2\times 6}$$

Take the square root of 25.

$$x=\frac{-\left(-1\right)±5}{2\times 6}$$

The opposite of -1 is 1.

$$x=\frac{1±5}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{1±5}{12}$$

Now solve the equation x=\frac{1±5}{12} when ± is plus. Add 1 to 5.

$$x=\frac{6}{12}$$

Reduce the fraction \frac{6}{12} to lowest terms by extracting and canceling out 6.

$$x=\frac{1}{2}$$

Now solve the equation x=\frac{1±5}{12} when ± is minus. Subtract 5 from 1.

$$x=-\frac{4}{12}$$

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

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

The equation is now solved.

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

Steps for Completing the Square

Subtract x from both sides.

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

Divide both sides by 6.

$$\frac{6x^{2}-x}{6}=\frac{1}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}-\frac{1}{6}x=\frac{1}{6}$$

Divide -\frac{1}{6}, the coefficient of the x term, by 2 to get -\frac{1}{12}. Then add the square of -\frac{1}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{6}x+\left(-\frac{1}{12}\right)^{2}=\frac{1}{6}+\left(-\frac{1}{12}\right)^{2}$$

Square -\frac{1}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{6}x+\frac{1}{144}=\frac{1}{6}+\frac{1}{144}$$

Add \frac{1}{6} to \frac{1}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{6}x+\frac{1}{144}=\frac{25}{144}$$

Factor x^{2}-\frac{1}{6}x+\frac{1}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{12}\right)^{2}=\frac{25}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{12}\right)^{2}}=\sqrt{\frac{25}{144}}$$

Simplify.

$$x-\frac{1}{12}=\frac{5}{12} x-\frac{1}{12}=-\frac{5}{12}$$

Add \frac{1}{12} to both sides of the equation.

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