Solve 1-1/x=-6x

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

Solve for x

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

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

Steps Using Factoring By Grouping

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$x-1=-6xx$$

Multiply x and x to get x^{2}.

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

Add 6x^{2} to both sides.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

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 positive, the positive number has greater absolute value than the negative. 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.

$$a=-2 b=3$$

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

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

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

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

Factor out common term 3x-1 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$x-1=-6xx$$

Multiply x and x to get x^{2}.

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

Add 6x^{2} to both sides.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

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}.

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

Square 1.

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

Multiply -4 times 6.

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

Multiply -24 times -1.

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

Add 1 to 24.

$$x=\frac{-1±\sqrt{25}}{2\times 6}$$

Take the square root of 25.

$$x=\frac{-1±5}{2\times 6}$$

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.

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

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

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

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

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

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

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

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$x-1=-6xx$$

Multiply x and x to get x^{2}.

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

Add 6x^{2} to both sides.

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

Add 1 to both sides. Anything plus zero gives itself.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

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.

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

Subtract \frac{1}{12} from both sides of the equation.

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